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. We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, so there isproving the existence of a proper class of inaccessible cardinals.
To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, ifgiven $\zeta < \kappa$ thenwe take a subset $z$ of $y$ of size $\zeta$. We have $z \in y$, so because $\mathcal{P}(\mathcal{P}(z)) \in y$$y$ contains its small subsets. So for every $A \in \mathcal{P}(\mathcal{P}(z))$ Then we have $\lbrace A\rbrace \in y$$\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation. Therefore Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements. This shows that $2^{2^{\zeta}} \le \kappa$, so and therefore that $2^{\zeta} < \kappa$.
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 becausecontradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)