If $R_{k} $ is a model for replacement, is necesarly k a strong limit cardinal?
And if k is a regular cardinal, are then equivalent the sentences:
k is a strongly inaccesible cardinal if and only if $ ZFC ^ {R_{K}} $ ? is an equivalence, or only one way of the biconditional is valid?
Note that if $R_\kappa$ is a model of the axiom of replacement as well the axiom of choice, then every set has an injection into the ordinals of $R_\kappa$, which is exactly $\kappa$. In particular, if $x\in R_\kappa$, $|\mathcal P(x)|\leq\kappa$.
This immediately tells us that $\kappa$ is a strong limit cardinal (and in fact that $\kappa=\beth_\kappa$). And strong limit cardinals are strongly inaccessible if and only if they are regular.
However it is not necessary, in general, that $\kappa$ is regular. Indeed, if such $\kappa$ exists that $R_\kappa$ is a model of replacement (and therefore of $\sf ZFC$), then the least such $\kappa$ is singular.
So the equivalence, in general, is false. But if one requires that $\kappa$ is regular it holds.