Answer
$\dfrac{3a^2b^2c^2}{4}$
Work Step by Step
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
$div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=\dfrac{\partial (x^2yz)}{\partial x}+\dfrac{\partial (xy^2z)}{\partial y}+\dfrac{\partial (xyz^2)}{\partial z}=6xyz$
Consider $I=\int_{0}^a\int_0^b \int_0^c (6xyz) \times dzdydx=\int_{0}^a\int_0^b [6xy(\dfrac{z^2}{2})]_0^c \times dydx$
Hence, we have $I=\int_{0}^a\int_0^b 3xyc^2 dydx=\dfrac{3a^2b^2c^2}{4}$
