I want to understand what the question asks about.
Question is,
Let $p$ be a prime integer. Find the smallest integer $n$ such that $F_{p^n}$ contains two subfields isomorphic to $F_{p^r}$ and $F_{p^s}$. (F means 'field'. I don't know how to represent this symbol in here..)
My answer is, if $F_{p^m}$ ⊂ $F_{p^n}$, then $n$ is divided by $m$. So, in same manner, $F_{p^r}$ ⊂ $F_{p^n}$ and $F_{p^s}$ ⊂ $F_{p^n}$, $n$ is divisible by $r$ and $s$. Thus, smallest integer $n$ is $\operatorname{lcd}(r,s)$.
But, I confuse about what's the meaning of 'isomorphic to $F_{p^r}$ and $F_{p^s}$'. $F_{p^n}$ ($n=\operatorname{lcd}(r,s)$) can be isomorphic to $F_{p^r}$ and $F_{p^s}$'?
I want to get an answer about this, and furthermore, the solution of this question!
Thank you for read this!
As Michael Klyachman pointed out, you are confusing with '$\mathbb{F}_{p^n}$ is isomorphic to $\mathbb{F}_{p^r}$' and '$\mathbb{F}_{p^n}$ contains an isomorphic copy of $\mathbb{F}_{p^r}$.' For example, the abelian group $\mathbb{Z}/4\mathbb{Z}$ contains a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$ (namely, the subgroup $\{\overline{0},\overline{2}\}$), but it does not mean $\mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
The problem asks when $\mathbb{F}_{p^n}$ can contain subfields isomorphic to $\mathbb{F}_{p^r}$ and $\mathbb{F}_{p^s}$ respectively. That is, $\mathbb{F}_{p^n}$ contains two subfields $K$ and $L$ such that $K\cong \mathbb{F}_{p^r}$ and $L \cong \mathbb{F}_{p^s}$.
You may know that $\mathbb{F}_{p^r}$ is a subfield of $\mathbb{F}_{p^n}$ if and only if $r\mid n$. The left-to-right direction is not hard to prove: think $\mathbb{F}_{p^n}$ as a $\mathbb{F}_{p^r}$-vector space and think about a basis $B$, and evaluate the size of $\mathbb{F}_{p^n}$ in terms of $\mathbb{F}_{p^r}$ and the size of $B$. You may refer to the remaining part of the proof from a previous answer on MSE.
Hence we have the solution: if $\mathbb{F}_{p^n}$ contains the isomorphic copy of $\mathbb{F}_{p^r}$ and $\mathbb{F}_{p^s}$, then both $r\mid n$ and $s\mid n$. Hence $\operatorname{lcd}(r,s)\mid n$. If $\operatorname{lcd}(r,s)=n$, then both of $r\mid n$ and $s\mid n$ hold, so $\mathbb{F}_{p^n}$ contains subfields that are isomorphic to $\mathbb{F}_{p^r}$ and $\mathbb{F}_{p^s}$ respectively.