Answer
$$\dfrac{2 \pi a^4}{3}$$
Work Step by Step
$\vec{r_u} \times \vec{r_v}=a\cos u \cos v i+a \sin u \cos v j -a \sin v k$
Now, $ |\vec{r_u} \times \vec{r_v}| =\sqrt {(a\cos u \cos v)^2 +(a \sin u \cos v)^2 +( -a \sin v )^2} \\ =\sqrt {a^4 \sin^2 v} \\=a^2 \sin v$
$$I=\int_{0}^{\pi/2} \int_{0}^{2 \pi} (a^2 \cos^2 v) (a^2 \sin v) \ du \ dv \\=a^4 \int_{0}^{\pi/2} \sin v \cos^2 v \times [u]_0^{2 \pi} dv \\=-2 \pi a^4 \times \times \dfrac{1}{3} [\cos^2 v]_{0}^{\pi/2} \\=\dfrac{-2 \pi a^4}{3} \times (\cos^3 \dfrac{\pi}{2}-\cos^3 0) \\=\dfrac{2 \pi a^4}{3}$$
