From Ackermann's set theory equals ZF (1970) by William N. Reinhardt:
Let A be the theory determined by the following axioms:
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x = y$
Class construction: $\exists x \forall y (y \in x \leftrightarrow y \in V \land \phi)$ where $x$ is not free in $\phi$
Strong completeness of V: $(y \in x \lor y \subseteq x) \land x \in V \to y \in V$
Ackermann's schema: $x, y \in V \to (\forall z)(\phi(x,y,z) \to z \in V) \to (\exists w \in V) (\forall z) (z \in w \leftrightarrow \phi(x,y,z))$ where $\phi$ does not mention $V$ and its only free variables are $x,y,z$
On page 191, the paper says
In the paper introducing the theory A [1], Ackermann showed that the relativizations to V of all the axioms of Zermelo's theory (Z) (except for the regularity axiom) could be proved in A. He also stated that the replacement schema of ZF (relativized to V) can be proved in A. An error in Ackermann's proof of this assertion was found by Levy [8], however, and the question remained open.
On page 192-193, it says
Recall that Ackermann originally stated that the replacement schema of ZF (relativized to V) can be proved in A. Notice that (2) establishes this with A replaced by A*, i.e. assuming the axiom of regularity for sets. Since A* is consistent relative to A in much the same way that ZF is consistent relative to ZF without regularity, this shows that Ackermann's statement is correct in the universe of "regular" sets and classes. I do not know whether the assumption of regularity is necessary.
Does A alone prove replacement (relativized to V)?