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It seems to be trivial: take $V_A:=A$.

EDIT: as it has been reformulated, the question has to be answered by "no" negatively now.

If you require $A$ and $V_A$ to have the same dimension $n$, and you ask that there exists a triple $(x,y,z)$ in $V_A$ such that $(xy)z\neq x(yz)$, then $N(V_A)\subsetneq V_A$ and thus $dim(N(V_A))<n$. So there is no hope to have $A=N(V_A)$ even at the level of vector spaces.

2 added 363 characters in body; edited body; added 6 characters in body

It seems to be trivial: take $V_A:=A$.

EDIT: as it has been reformulated, the question has to be answered by "no" now.

If you require $A$ and $V_A$ to have the same dimension $n$, and you ask that there exists a triple $(x,y,z)$ in $V_A$ such that $(xy)z\neq x(yz)$, then $N(V_A)\subsetneq V_A$ and thus $dim(N(V_A))<n$. So there is no hope to have $A=N(V_A)$ even at the level of vector spaces.

1

It seems to be trivial: take $V_A:=A$.