*Really, this was answered in the comments; I'm putting this answer down to move this off the unanswered queue. I've made this CW and will delete it if one of the original commenters adds their own answer.*

We have in $L$, for each (infinite) $\alpha$, the following bijections:

Hence $\vert\mathcal{P}(L_\alpha)^L\vert=\vert(\vert\alpha\vert^+)^L\vert$. Now assuming $0^\sharp$ we have that $\omega_1^V$ is a limit cardinal in $L$, so for each $\alpha<\omega_1^V$ we have $\vert\mathcal{P}(L_\alpha)^L\vert=\aleph_0$.

So the answer to your question is $\omega_1^V$.

Note that all this requires is that $\omega_1^V$ be a limit cardinal in $L$. More generally, let $\kappa$ be the supremum of the $L$-cardinals whose $L$-successor is $<\omega_1^V$; then the $\kappa$th level of $L$ is the first whose $L$-powerset is truly uncountable.