The question makes the most sense if we assume $V=L$, since otherwise it could be that every set of reals in $L$ is countable and hence measurable. 

If $V=L$, the answer is $\omega_1+1$. Every set in $L_{\omega_1^L}$ is countable in $L$ and hence measurable, and so it cannot be $\omega_1$ or less. But at stage $\omega_1$, the set $2^\omega$ is definable in $L_{\omega_1}$, and we can also define the $L$-order on these reals. So we can define a Vitali set at this level. Namely, we can define the tail-equivalence relation on binary sequences, eventual agreement, and I can pick the $L$-least member of each equivalence class, all definably over $L_{\omega_1^L}$. This set is not measurable using the coin-flipping probability measure on $2^\omega$ in $L$, since we can consider the sets that arise by flipping finitely many digits. These are disjoint, for all the countably many ways to flip digits differently, but the resulting sets all have the same measure as the original, and they cover $2^\omega$, contradiction.

So the nonmeasurable sets show up at the same time as the whole set $2^\omega$ itself, namely, in $L_{\omega_1+1}$.