 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 givesIf $a$ is empty, $\alpha =0$ works. No assumption that $a$ itself belongs to $L$ is used; it is enough that its elements do.