Answer
$(\sqrt 3 -1) \dfrac{1}{3} \pi a^3$
Work Step by Step
Apply the spherical coordinates system as:
$x=\rho \sin \phi \cos \theta \\ y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi$;
So, $\rho=\sqrt {x^2+y^2+z^2} \implies \rho^2=x^2+y^2+z^2$
The jacobian for spherical coordinates is $\rho^2 \sin \rho$, therefore, $\iiint{E} \sqrt {x^2+y^2+z^2} dV =\iiint_{E} \rho^2 \sin \rho d\rho d\theta d\phi$
Now, we will consider $I=\int_{\pi/6}^{\pi/3} \int_{0}^{2\pi}\int_{0}^{a} \rho^2 \sin \phi d\rho d\theta d\phi$
$=\int_{\pi/6}^{\pi/3} \int_{0}^{2\pi}[\dfrac{1}{3} \rho^3 \sin \phi ]_0^a d\theta d\phi \\ = \int_{\pi/6}^{\pi/3} \int_{0}^{2\pi} \dfrac{1}{3} a^3 \sin \phi d\theta d \phi \\=\dfrac{2}{3} [-\cos\phi \pi a^3]_{\pi/6}^{\pi/3}$
Hence, $I=\dfrac{1}{3} \pi a^3-\dfrac{1}{3} \pi a^3 =(\sqrt 3 -1) \dfrac{1}{3} \pi a^3$
