Answer
$\dfrac{\pi}{8}$
Work Step by Step
$I=\int_0^{\pi/2} \int_{0}^{\pi/2}\int_{0}^{1} (\rho \sin \phi \cos \theta) e^{\rho^2} (\rho^2 \sin \phi) d\rho d\theta d\phi$
or, $I=\int_0^{\pi/2} ( \sin^2 \phi d\phi) \times \int_{0}^{\pi/2} \cos \theta d\theta \times \int_{0}^{1} \rho^3 e^{\rho^2} d\rho $
or, $I=[\dfrac{\phi}{2}-\dfrac{\sin 2\phi}{4}]_0^{\pi/2} \times (1-0) \times [ \dfrac{1}{2} e^{(1)^2} (1-1)- \dfrac{1}{2} e^{(0)^2} (0^2-1)]$
or, $I=\dfrac{\pi}{4}[0-\dfrac{1}{2}(-1)]$
Thus, $=\dfrac{\pi}{8}$
