Yes, the toral component of a connected Lie group is equal to the toral component of its solvable radical.

Let $G$ be a Lie group, $S$ its solvable radical, and $\mathrm{TC}(G)$ denote the toral component of $G$. As abx has noticed, $\mathrm{TC}(G)\subseteq S$, hence
$\mathrm{TC}(G)\subseteq\mathrm{TC}(S)$. We show that
$\mathrm{TC}(S)\subseteq\mathrm{TC}(G)$.

Since $S$ is a characteristic subgroup of $G$, we see that $\mathrm{TC}(S)$ is a characteristic subgroup of $G$, hence it is a normal subgroup of $G$, hence $G$
acts on $\mathrm{TC}(S)$ by conjugation. Since $T:=\mathrm{TC}(S)$ is a torus in the sense of Lie group theory, we have $T\simeq (\mathbb{R}/\mathbb{Z})^n$ for some natural $n$, hence $\mathrm{Aut}(T)\simeq\mathrm{GL}(n,\mathbb{Z})$, hence the connected component of the identity in $\mathrm{Aut}(T)$ is trivial. It follows that the connected Lie group $G$ acts *trivially* on $T$ when acting by conjugation. This means that $T=\mathrm{TC}(S)$ is contained in the center of $G$. Thus $\mathrm{TC}(S)\subseteq\mathrm{TC}(G)$, as required.