Let $R_\alpha$ be the $\alpha$-th infinite regular ordinal. This question assumes AC, so the following is true:

$$R_\alpha=\left\{ \begin{array}{ll} \omega_\alpha & \alpha=\kappa+n\;\mathrm{where}\;n<\omega\land\kappa\;\mathrm{is}\;\mathrm{wk. inaccessible}\;\mathrm{or}\;0 \\ \omega_{\alpha+1} & \mathrm{otherwise} \end{array} \right.$$

After this point, **one could define an $\alpha$-Supercorrect cardinal $\kappa$ as follows:**

$$V_\kappa\prec_{L_{R_\alpha,R_\alpha}}V$$

Here are some remarks:

- The $0$-Supercorrect cardinals are precisely the Correct cardinals (which are equiconsistent with ZFC).
- In the rank of any $\alpha$-Supercorrect cardinal, there are $\beta$-Supercorrect cardinals for every $\beta<\alpha$. (Thus, the $\alpha$-Supercorrect cardinals are stronger in consistency strength the greater the $\alpha$)
- Every $\alpha$-Supercorrect cardinal is $\beta$-Supercorrect for every $\beta<\alpha$
- Every $1$-Supercorrect $\kappa$ is Worldly, Correct, and $V_\kappa$ satisfies the existence of a Correct cardinal.
- In the rank of a $\kappa$ which is $\kappa$-Supercorrect, there is a proper class of $\beta$-Supercorrects for every $\beta$. Of course, this means that the consistency strength of a $\kappa$ which is $\kappa$-Supercorrect is stronger than that of "For every $\beta$ there is a proper class of $\beta$-Supercorrect cardinals".
- For every first-order theory $T$, a $1$-Supercorrect cardinal has $V_\kappa\models T\Leftrightarrow V\models T$.

**Are any of these inconsistent? Are all other than $0$-Supercorrect inconsistent? What else would be true if consistent?**

*Edit:*

Interesting result found: it is inconsistent for there to be a cardinal $\kappa$ which is $\alpha$-Supercorrect for every $\alpha$. If there were, it turns out that there would be another such cardinal in $V_\kappa$.

$L_{\infty,\infty}$ can express all of its own schemas as singular formulae. Therefore, one could let the schema $V_\kappa\prec_{L_{\infty,\infty}}V$ be the union of "$V_\kappa\models\phi\Leftrightarrow\phi$" for every $L_{\infty,\infty}$ $\phi$. Since this is expressable as a singular formula in $L_{\infty,\infty}$, we could call this singular formula $\psi(\kappa)$. If there is such a $\kappa$, $V\models\exists\kappa(\psi(\kappa))$. Therefore, $V_\kappa\models\exists\kappa(\psi(\kappa))$.