Answer
There does not exist such a vector field $G$.
Work Step by Step
Suppose, we have a vector field $G$ such that $div [curl (G)]=0$
when $F=A i+B j+C k$, then we have $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$
Given: $curl G=xyz i-y^2 zj+yz^2 k$
$div[curl(G)]=div (xyz i-y^2 zj+yz^2 k) $
and $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}=\dfrac{\partial xyz}{\partial x}+\dfrac{\partial (-y^2z)}{\partial y}+\dfrac{\partial (yz^2)}{\partial z}$
$\implies div[curl(G)]=yz-2yz+2yz=yz$
Hence, we can conclude that there does not exist such a vector field $G$.
