Answer
$-8 \pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
or, $\oint F \cdot dr=\int_{0}^{2 \pi} \int_{0}^{2} (-3x^2y^2) dy dx$
or, $\int_{0}^{2 \pi} \int_{0}^{2} (-3(r\cos \theta)^2(r \sin \theta)^2) dr d\theta=(-32) \int_{0}^{2 \pi} \int_{0}^{2} (\sin \theta \cos \theta)^2dr d\theta $
This implies that
$(-8)\int_{0}^{2 \pi} \int_{0}^{2}(\dfrac{1}{2}) (1-\cos 4\theta)d\theta =-8 \pi$
