The correct answer is $\boxed{\left( {4{\rm{y}}{{\rm{z}}^3} + 2{\rm{x}}{{\rm{y}}^2}} \right){\rm{\hat i}} + 2{{\rm{x}}^2}{\rm{z\hat j}} – 2{{\rm{y}}^2}{\rm{z\hat k}}}$.
The curl of a vector field $\overrightarrow F$ is defined as
$$\nabla \times \overrightarrow F = \left( \dfrac{\partial Q}{\partial x} – \dfrac{\partial P}{\partial z} \right){\rm{\hat i}} + \left( \dfrac{\partial R}{\partial y} – \dfrac{\partial Q}{\partial z} \right){\rm{\hat j}} + \left( \dfrac{\partial P}{\partial y} – \dfrac{\partial R}{\partial x} \right){\rm{\hat k}}$$
where $P$, $Q$, and $R$ are the $x$, $y$, and $z$ components of $\overrightarrow F$, respectively.
In this case, $\overrightarrow F = {{\rm{x}}^2}{{\rm{z}}^2}{\rm{\hat i}} – 2{\rm{x}}{{\rm{y}}^2}{\rm{z\hat j}} + 2{{\rm{y}}^2}{{\rm{z}}^3}{\rm{\hat k}}$. Therefore,
$$P = {{\rm{x}}^2}{{\rm{z}}^2}, \quad Q = -2{\rm{x}}{{\rm{y}}^2}, \quad R = 2{{\rm{y}}^2}{{\rm{z}}^3}$$
Substituting these values into the formula for the curl, we get
$$\nabla \times \overrightarrow F = \left( \dfrac{\partial}{\partial x} \left( 2{{\rm{y}}^2}{{\rm{z}}^3} \right) – \dfrac{\partial}{\partial z} \left( {{\rm{x}}^2}{{\rm{z}}^2} \right) \right){\rm{\hat i}} + \left( \dfrac{\partial}{\partial y} \left( -2{\rm{x}}{{\rm{y}}^2} \right) – \dfrac{\partial}{\partial z} \left( 2{{\rm{y}}^2}{{\rm{z}}^3} \right) \right){\rm{\hat j}} + \left( \dfrac{\partial}{\partial y} \left( {{\rm{x}}^2}{{\rm{z}}^2} \right) – \dfrac{\partial}{\partial x} \left( -2{\rm{x}}{{\rm{y}}^2} \right) \right){\rm{\hat k}}$$
$$= \left( 6{{\rm{y}}^2}{{\rm{z}}^2} – 2{{\rm{x}}^2}{{\rm{z}}^3} \right){\rm{\hat i}} + \left( -4{\rm{x}}{{\rm{y}}^3} – 2{{\rm{y}}^2}{{\rm{z}}^3} \right){\rm{\hat j}} + \left( 2{{\rm{x}}^2}{{\rm{z}}^2} + 4{\rm{x}}{{\rm{y}}^2} \right){\rm{\hat k}}$$
$$= \left( 4{\rm{y}}{{\rm{z}}^3} + 2{\rm{x}}{{\rm{y}}^2}} \right){\rm{\hat i}} + 2{{\rm{x}}^2}{\rm{z\hat j}} – 2{{\rm{y}}^2}{\rm{z\hat k}}$$
Therefore, the curl of $\overrightarrow F$ is $\boxed{\left( {4{\rm{y}}{{\rm{z}}^3} + 2{\rm{x}}{{\rm{y}}^2}} \right){\rm{\hat i}} + 2{{\rm{x}}^2}{\rm{z\hat j}} – 2{{\rm{y}}^2}{\rm{z\hat k}}}$.