Answer
$\dfrac{\pi}{8}$
Work Step by Step
In the spherical coordinates system, we have
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
Now, consider $I=\int_0^{\pi/2} \int_{0}^{\pi/2}\int_{0}^{1} \times (\rho \sin \phi \cos \theta) \times e^{\rho^2} \times (\rho^2 \sin \phi) d\rho d\theta d\phi$
$=\int_0^{\pi/2} ( \sin^2 \phi d\phi) \cdot \int_{0}^{\pi/2} \cos \theta d\theta \cdot \int_{0}^{1} \rho^3 e^{\rho^2} d\rho $
Hence, $I=[\dfrac{\phi}{2}-\dfrac{\sin 2\phi}{4}]_0^{\pi/2} [\sin (\pi/2)-\sin 0] \cdot [ \dfrac{1}{2} e^{(1)^2} (1-1)- \dfrac{1}{2} e^{(0)^2} (0^2-1)]=\dfrac{\pi}{8}$
