Solution
= Solution
For each $x\in a$, let $\beta_x$ be the least <ordinal> with $x\in L_{\beta_x}$. It exists because $a\subseteq L$, and it is uniquely definable. <Replacement> therefore gives the set $\{\beta_x:x\in a\}$. Choose an <ordinal> $\alpha$ at least their <supremum>. Increasingness of the hierarchy gives
$$
\boxed{a\subseteq L_\alpha.}
$$
If $a$ is empty, $\alpha=0$ works. No assumption that $a$ itself belongs to $L$ is used; it is enough that its elements do.