**Claim:** $|V_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$ implies that $0^\#$ exists.

**Proof:** Let's assume, toward contradiction, that $0^\#$ doesn't exist that $|V_\alpha \cap L| = |\alpha|$ for every $\alpha \geq \omega$.

Let $\mu$ be a singular cardinal in $V$. Let's apply the assumption $|V_\alpha \cap L| = |\alpha|$ for $\alpha = \mu + 1$. In $V$, $\mu = |\mu + 1|=|V_{\mu + 1} \cap L| \geq |\mathcal{P}^L(\mu)| \geq |(\mu^{+})^L|$.

In particular, $\mu^{+} > (\mu^{+})^L$. Therefore, the cofinality of $(\mu^{+})^L$ in $V$ is strictly below $\mu$. Applying Jensen's covering theorem, we conclude that the cofinality of $(\mu^{+})^L$ in $L$ is strictly below $\mu$ - a contradiction to the regularity of $(\mu^{+})^L$ in $L$.

By the way, if we restrict the values of $\alpha$ for which we want $|V_\alpha \cap L| = |\alpha|$ to be $\leq \aleph_{\omega}$, the consistency strength drops considerably:

**Claim:** $|V_\alpha \cap L| = |\alpha|$ for every $\omega \leq \alpha \leq \aleph_\omega$ is equiconsistent with the existence of $\omega$ inaccessible cardinals.

**Proof:** For the first direction, force over $L$ with an iteration of Levi collapses in order to make the $n$-th inaccessible in $L$ equal to the $\aleph_n$ of the generic extension. For the second direction, note that the assumption implies that for all $1\leq n < \omega$, $\aleph_n$ is a $\beth$-fixed point in $L$, since for every infinite $\beta < \aleph_n$, $|V_\beta \cap L| = |\beth_\beta^L| < \aleph_n$. In particular, $\aleph_n$ is a limit cardinal in $L$. It is regular in $V$ and thus also in $L$ and therefore it is inaccessible.