Answer
There is no vector field $G$.
Work Step by Step
Use definition of div $F=\nabla \cdot F$
Consider $F=ai+bj+ck$
That is, div $F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$
The div (curl G) will be zero when there exists a vector field $G$ for all values of $x,y$ and $z$.
Here, we have curl$G=2x i+3y z j-xz^2 k$
Thus, div (curl G)$=\dfrac{\partial (2x)}{\partial x}+\dfrac{\partial (3yz)}{\partial y}+\dfrac{\partial (-xz^2)}{\partial z}=-2+3z-2xz$
We can see that div (curl G) is not zero, so there is no such vector field $G$.
Hence, it has been proved.
