I know that in the product (Tychonoff) topology equality $${\rm int}\left(\prod_{\alpha \in I} A_ {\alpha}\right)= \prod_{\alpha \in I} {\rm int}\left( A_ {\alpha}\right)$$ does not hold true (see question Interior of arbitrary products).
Does above equality hold in case the product space is given the box topology?
Yes, it does hold.
Note that since each $\DeclareMathOperator{\Int}{Int}\Int (A_\alpha)$ is open in $X_\alpha$ it follows that $\prod_{\alpha \in I} \Int ( A_\alpha )$ is open in $\Box_{\alpha \in I} X_\alpha$, and is clearly a subset of $\prod_{\alpha \in I} A_\alpha$. It thus follows that $${\textstyle \prod_{\alpha \in I}} \Int ( A_\alpha ) \subseteq \Int ( {\textstyle \prod_{\alpha \in I}} A_\alpha ).$$
Now let's take some $\langle x_\alpha \rangle_{\alpha \in I} \in \Int ( \prod_{\alpha \in I} A_\alpha )$. But as $\Int ( \prod_{\alpha \in I} A_\alpha )$ is open, this means that for each $\alpha \in I$ there is an open $U_\alpha \subseteq X_\alpha$ containing $x_\alpha$ such that $${\textstyle \prod_{\alpha \in I}} U_\alpha \subseteq \Int ( {\textstyle \prod_{\alpha \in I}} A_\alpha ) \subseteq {\textstyle \prod_{\alpha \in I}} A_\alpha.$$ But in order for this to be true we must have that $U_\alpha \subseteq A_\alpha$ for all $\alpha$, which implies $U_\alpha \subseteq \Int ( A_\alpha)$. Therefore $\langle x_\alpha \rangle_{\alpha \in I} \in \prod_{\alpha \in I} \Int ( A_\alpha )$.
