Here is Prob. 18, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Let $E$ have finite outer measure. Show that there is an $F_\sigma$ set $F$ and a $G_\delta$ set $G$ such that $F \subseteq E \subseteq G$ and $m^*(F) = m^*(E) = m^*(G)$.
My Attempt:
Case 1. If set $E$ is a (Lebesgue) measurable set of real numbers, then by Theorem 11 (iv), Chap. 2, in Royden, there is an $F_\sigma$ set $F$ with $F \subseteq E$ and $m^*(E \setminus F) = 0$, and by Theorem 11 (ii), there is a $G_\delta$ set $G$ such that $E \subseteq G$ and $m^*(G \setminus E) = 0$.
Note that the sets $F$ and $G$ both are measurable.
Now since $E$ is measurable with finite outer measure and since $F \subseteq E$, therefore $F$ also has finite outer measure.
Thus we have $F \subseteq E \subseteq G$ and $m^*(E \setminus F) = 0$ and $m^*(G \setminus E) = 0$, which by the excision property of measurable sets (i.e. sets $F$ and $E$) with finite outer measure yield $m^*(E) - m^*(F) = 0$ and $m^*(G) - m^*(E) = 0$, and thus $$ m^*(F) = m^*(E) = m^*(G), $$ as required.
Is what I have done correct and accurate in each and every detail? If so, then how to tackle the case when $E$ has finite outer measure but is not measurable?
PS:
Case 2. Now suppose that set $E$ has finite outer measure but is not measurable. Then by Theorem 11 (ii) and (iv) in Royden, for any $F_\sigma$ set $F$ and any $G_\delta$ set $G$ with $F \subseteq E \subseteq G$, we must have $m^*(E \setminus F) \neq 0$ and $m^*(G \setminus E) \neq 0$, that is, $m^*(E \setminus F) > 0$ and $m^*(G \setminus E) > 0$, which imply $E \setminus F \neq \emptyset$ and $G \setminus E \neq \emptyset$; in fact the sets $E \setminus F$ and $G \setminus E$ cannot even be countable.
Now as $E = F \cup (E \setminus F)$, so we have $$ m^*(E) = m^* \big( F \cup (E \setminus F) \big) \leq m^*(F) + m^*( E \setminus F), $$ which implies $$ m^*(E) - m^*(F) \leq m^*( E \setminus F), $$ but since $F \subseteq E$, by the monotonicity of the outer measure we have $$ 0 \leq m^*(E) - m^*(F) \leq m^*( E \setminus F). \tag{1} $$
And, as $G = E \cup (G \setminus E)$, so we have $$ m^*(G) = m^* \big( E \cup (G \setminus E) \big) \leq m^*(E) + m^*( G \setminus E ), $$ which implies $$ m^*(G) - m^*(E) \leq m^*(G \setminus E), $$ but since $E \subset G$, by the monotonicity of the outer measure we have $$ 0 \leq m^*(G) - m^*(E) \leq m^*(G \setminus E). \tag{2} $$ The relations (1) and (2) hold for any $F_\sigma$ set $F$ and for any $G_\delta$ set $G$ such that $F \subseteq E \subseteq G$.
What next? Where can we get from here?
PS (Based on the comments by Tab1e):
As set $E$ has finite outer measure, so by the definition of the outer measure, for each positive integer $n$, we can find a countable collection $\left\{ I_{n, k} \right\}_{k=1}^\infty$ of non-empty bounded open intervals covering set $E$ for which $$ \sum_{k=1}^\infty l \left( I_{n, k} \right) < m^*(E) + \frac{1}{n}, $$ Let us put $$ G_n := \bigcup_{k=1}^\infty I_{n, k}. $$ Then by our choice of the $I_{n, k}$, we have $E \subset G_n$, and also $$ m^*\left( G_n \right) = m^* \left( \bigcup_{k=1}^\infty I_{n, k} \right) \leq \sum_{k=1}^\infty m^* \left( I_{n, k} \right) = \sum_{k=1}^\infty l \left( I_{n, k} \right) < m^*(E) + \frac{1}{n}, $$ and hence we also have $$ m^* \left( G_n \right) < m^*(E) + \frac{1}{n}. $$
Let us now put $$ G := \bigcap_{n=1}^\infty G_n. $$ This set $G$ is of course a $G_\delta$ set, and simce for each positive integer $n$ we have $G \subset G_n$ and $E \subset G_n$, we can conclude that $E \subset G$ also and therefore $$ m^*(E) \leq m^* (G) \leq m^* \left( G_n \right) < m^*(E) + \frac{1}{n}, $$ which in turn implies that $$ m^*(E) \leq m^*(G) < m^*(E) + \frac1n $$ for every positive integer $n$, which upon taking the limit as $n \to \infty$ yields $$ m^*(E) = m^*(G). $$
Is this part of my post correct? If so, then how to give the proof for the $F_\sigma$ set?
The way you find sets $F$ and $G$ in your posting under the assumption that $E$ is measurable (in the sense of Caratheodory) is correct. The problem is with the statement of the problem. I check the statement in the book you quote and it seems surviving author might have missed the measurable adjective for $E$, whether that change in subsequent revisions, I don't know.
Here we show that if $E$ satisfies the conclussion of the problem, then $E$ must be measurable in the sense of Caratheodory). By Caratheodory's theorem, the outer measure $m^*$ extends $m$ to a measure on a $\sigma$-algebra $\mathcal{M}$ that contains the Borel $\sigma$-algebra $\mathscr{B}(\mathbb{R})$ as well as the $m$-null sets (sets with outer measure $0$).
If $F$ is an $F_\sigma$ set and $G$ is a $G_\delta$ set, then both are Borel measurable. So, if in addition, $F\subset E\subset G$, $m^*(F)=m^*(G)=m^*(E)<\infty$, then $m^*(G\setminus F)=0$. Indeed, since $F$ and $G$ are Borel sets $$\mu(G)=\mu^*(G)=m^*(F)+m^*(G\setminus F)=m(F)+m(G\setminus F)$$ Since $m^*(E)<\infty$, $m(G\setminus F)=m(G)-m(F)=0$. This would imply the $m^*$-measurability of $E$ for $$E=F\cup(E\setminus F),$$ $F\in\mathscr{B}(\mathbb{R})\subset\mathcal{M}$, and $E\setminus F\in\mathcal{M}$ on account that $m^*(E\setminus F)\leq m(G\setminus F)=0$.
As a counter example to the statement of the problem as it is written in the book, consider a the classical Vitali set $A$ constructed using the Axiom of choice and the equivalence $r\sim s$ if $r-s\in\mathbb{Q}$. $A$ set is not Lebesgue measurable measurable; $E=A\cap[0,1]$ satisfies $m^*(E)=1$ and $$\sup\{m(F):F\,\text{closed}\,,F\subset E\}=0$$