A cardinal $\kappa$ is $\Sigma_2$ reflecting if whenever a sentence $\varphi$ is true in some $V_\alpha$, then there is an $\alpha\lt\kappa$ such that $V_\alpha\models\varphi$.
(This can be seen to be equivalent to the "reflecting" version of $\Sigma_2$-reflecting, since every $\Sigma_2$ statement $\psi$ is equivalent to a statement of the form "$\exists\alpha V_\alpha\models\varphi$", where $\varphi$ is some sentence (with no bound on the complexity of $\varphi$). But let us leave this aside.)
Now, if $\kappa$ is a strong cardinal, and there is some $\alpha$ above $\kappa$ such that $V_\alpha\models\varphi$ for some statement $\varphi$, then fix an $\alpha$-strongness embedding $j:V\to M$ with critical point $\kappa$. Since $V_\alpha^M=V_\alpha$, the model $M$ thinks there is $\alpha\lt j(\kappa)$ for which $V_\alpha\models\varphi$. Thus, by elementarity, there is $\alpha\lt\kappa$ in $V$ for which $V_\alpha\models \varphi$, thereby verifying that $\kappa$ is $\Sigma_2$-reflecting, as desired.
Now, to answer your question, pick any $\gamma\gt\kappa$ such that all $\Sigma_2$ statements reflect from $V$ down to $V_\gamma$. Such a cardinal $\gamma$ exists by the Reflection theorem. Now, I claim that any $\gamma$-strong cardinal $\kappa$ is $\Sigma_2$ reflecting, since any statement $\varphi$ true in some $V_\alpha$ will be true in some $V_\alpha$ with $\alpha\lt\gamma$ by the choice of $\gamma$, and thus $\varphi$ will be true in $V_\alpha$ for some $\alpha\lt\kappa$ by the argument of the previous paragraph.
In fact, it is easy to see that if $\kappa$ is $\gamma$-strong for this $\gamma$, then in fact it is fully strong, since any violation of strongness of $\kappa$ above $\gamma$ would be reflected below $\gamma$.