Answer
$$\frac{{\left( {1 - {e^{ - 8}}} \right){\pi ^2}}}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^\pi {\int_0^2 {{e^{ - {\rho ^3}}}{\rho ^2}} d\rho } d\theta } d\phi \cr
& = \int_0^{\pi /2} {\int_0^\pi {\left[ {\int_0^2 {{e^{ - {\rho ^3}}}{\rho ^2}} d\rho } \right]} d\theta } d\phi \cr
& {\text{Rewrite}} \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\int_0^\pi {\left[ {\int_0^2 {{e^{ - {\rho ^3}}}\left( { - 3{\rho ^2}} \right)} d\rho } \right]} d\theta } d\phi \cr
& {\text{Integrate with respect to }}\rho \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\int_0^\pi {\left[ {{e^{ - {\rho ^3}}}} \right]_0^2} d\theta } d\phi \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\int_0^\pi {\left( {{e^{ - 8}} - {e^0}} \right)} d\theta } d\phi \cr
& = - \frac{1}{3}\int_0^{\pi /2} {\left[ {\int_0^\pi {\left( {{e^{ - 8}} - 1} \right)} d\theta } \right]} d\phi \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\int_0^{\pi /2} {\left[ {\int_0^\pi {d\theta } } \right]} d\phi \cr
& {\text{Integrate with respect to }}\theta \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\int_0^{\pi /2} {\left[ \theta \right]_0^\pi } d\phi \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\int_0^{\pi /2} \pi d\phi \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\pi \int_0^{\pi /2} {d\phi } \cr
& {\text{Integrate}} \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\pi \left[ \phi \right]_0^{\pi /2} \cr
& = - \frac{1}{3}\left( {{e^{ - 8}} - 1} \right)\pi \left( {\frac{\pi }{2}} \right) \cr
& = \frac{{\left( {1 - {e^{ - 8}}} \right){\pi ^2}}}{6} \cr} $$
