Yes.  Assume ZFC.  If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set.  Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$, and it's not hard to show that $|y|$ is an inaccessible cardinal greater than $|x|$, so there is a proper class of inaccessible cardinals.

To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, if $\zeta < \kappa$ then take a subset $z$ of $y$ of size $\zeta$.  We have $z \in y$, so $\mathcal{P}(\mathcal{P}(z)) \in y$.  So for every $A \in \mathcal{P}(\mathcal{P}(z))$ we have $\lbrace A\rbrace \in y$.  Therefore $2^{2^{\zeta}} \le \kappa$, so $2^{\zeta} < \kappa$.

To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, a contradiction because $\kappa^{cof( \kappa)} > \kappa$.