Answer
$-24 \pi$
Work Step by Step
Stoke's Theorem: Let us consider that $C$ be the closed curve which encloses surface $S$ for a vector field $F$. This also states that the line integral and the surface integral must be equal. That is,
$\oint_C F \ dx=\iint_S (\nabla \times F) \cdot n \ dS$
Here, $\nabla \times F=\begin{vmatrix} i&j&k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ 2y & -z & x \end{vmatrix}=(1,-1,-2)$ and $n=(0,0,1)$
Thus, the surface integral can be computed as:
$\iint_S (\nabla \times F) \cdot n \ dS=\iint_S (1,-1,-2) \cdot (0,0,1) \ dS\\=-2 \int_0^{2\pi} \int_0^{2 \sqrt 3} r \ dr \ d \theta\\=-2 \times 2 \pi \times 6\\=-24 \pi$
