Answer
$\dfrac{12 \pi a^5}{5}$
Work Step by Step
As we know that $div F=\dfrac{\partial A}{\partial x}i+\dfrac{\partial B}{\partial y}j+\dfrac{\partial C}{\partial z}k$
Now, we have
$Flux =\iiint_{o} 3x^2+3y^2+3z^2 dA$
Also,
$Flux =\nabla \cdot F=(3) \int_{0}^{2 \pi}\int_{0}^{\pi}\int_{0}^{a} (\rho^2) (\rho^2 \sin \phi) d\rho d \phi d\theta$
This implies that
$(3) \int_{0}^{2 \pi}(\dfrac{2a^5}{5} )d\theta = \dfrac{12 \pi a^5}{5}$
