Let $L=\mathfrak{sl}(V)$ (matrices with trace zero). Use Lie's theorem to prove that Rad $L = Z(L).$
This exercise is from Humphrey's and has the following hint:
[Observe that Rad $L$ lies in each maximal solvable subalgebra $B$ of $L$. Select a basis of $V$ so that $B=L\cap \mathfrak t(n,F)$ (upper triangular matrices) and notice that the transpose of $B$ is also a maximal solvable subalgebra of $L$. Conclude that Rad $L \subset L\cap \mathfrak d(n,F)$ (diagonal matrices), then that Rad $L = Z(L)]$
Following the hint, I would like to check how to justify all the steps correctly.
Since Rad $L$ is a solvable ideal, in particular it is also a solvable subalgebra. Hence it lies in each maximal solvable subalgebra $B$ of $L$. (I can't justify this, because inclusion is not a total order. Does it follows from Zorn's lemma? ) Now, using Lie's theorem we may find a basis for $V$ in which every matrix of $B$ is upper triangular, ie, every element of $B$ lies in $\mathfrak t(n,F).$ The transpose of $B$ is also a maximal subalgebra of $L$ (why? because $[x^t,y^t]= [yx]^t? $ and why would it still be maximal?). The matrices which lie in $B\cap B^t$ are all diagonal, and since $B,B^t$ are both maximal, we have Rad $L$ lying in this intersection. Therefore, Rad $L\subset L\cap\mathfrak d(n,F)$, the diagonal matrices. This implies Rad $L\subset Z(L) $ (why? we have $Z(L)$ as the scalar matrices in field of characteristic $n.$)
I assume that $V$ is an $n$-dimensional vector space over some field $F$ with $n\in\mathbb{Z}_{\ge0}$, so that $L$ is a finite-dimensional Lie algebra over $F$. Let $I:=\text{Rad}(L)$. Note that $B+I$ is a solvable subalgebra of $L$. Since $B$ is a maximal solvable subalgera, $B+I=B$, whence $I\subseteq B$.
Now, if the transpose $B^\top$ of $B$ is not a maximal, then $B^\top$ is contained in a larger solvable subalgebra $S$ of $L$. Then, $B$ is properly contained in the solvable subalgebra $S^\top$, yielding a contradiction.
As for why the center $Z(L)$ of $L$ contains only scalar matrices, you can see this link. On the other hand, if you are asking why $\text{Rad}(L)\subseteq L\cap \mathfrak{d}(n,F)$ implies that $\text{Rad}(L)\subseteq Z(L)$, then observe that $$\left[E_{i,j},E_{k,k}\right]=\delta_{j,k}\,E_{i,k}-\delta_{i,k}\,E_{k,j}$$ for all $i,j,k=1,2,\ldots,n$. Here, for each $i,j=1,2,\ldots,n$, $E_{i,j}\in\mathfrak{gl}(V)$ (not $\mathfrak{sl}(V)$) denotes the $n$-by-$n$ matrix with $1$ at the $(i,j)$-entry and $0$ everywhere else, and $\delta$ is the Kronecker delta. Thus, if we have $\sum\limits_{k=1}^n\,\alpha_k\,E_{k,k}\in \text{Rad}(L)$ for some $\alpha_1,\alpha_2,\ldots,\alpha_n\in F$, then $$\left[E_{i,j},\sum_{k=1}^n\,\alpha_k\,E_{k,k}\right]=\alpha_j\,E_{i,j}-\alpha_i\,E_{i,j}$$ for every $i,j=1,2,\ldots,n$. Since $\text{Rad}(L)$ is an ideal of $L$, we have $$\alpha_j\,E_{i,j}-\alpha_i\,E_{i,j}\in \text{Rad}(L)\subseteq \mathfrak{d}(n,F)\,,$$ whence $\alpha_j=\alpha_i$ for every $i,j=1,2,\ldots,n$ with $i\neq j$.