I'll be using $(\rho,\theta,\phi)$ to represent respectively the radius, azimuth angle (the one in the $xy$ plane) and zenith angle (the one measured from the $z$-axis). Also, since the problem is trivial when $a=0$, assume $a>0$.
First, notice that the sphere is symmetrical with respect to $\theta$, so we can fix a vertical cross section to establish our $\rho$ and $\phi$ integration limits. Here is a picture of the sphere in the $zx$-plane (sorry about the glare):

Now, notice right away that we must have $0\leq\phi\leq\pi/4$. $\phi=0$ represents the coordinate point $(0,0,2a)$, while $\phi=\pi/4$ represents the coordinate point $(a,0,a)$. If you're not quite convinced of this second one just draw a triangle. So the only difficult part really is establishing the $\rho$ limits, which will be in terms of $\phi$. In other words, our integral will look like: $ \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\pi/4}\int_{\rho=\rho_1(\phi)}^{\rho=\rho_2(\phi)}\rho^2\sin\phi d\rho d\phi d\theta $
I've labelled the $\rho$ limits $\rho_1$ and $\rho_2$ in the picture ($\rho_2$ is a little hard to see - its the distance from the origin to the point on the surface of the sphere).
For $\rho_1$: the triangle drawn illustrates how to find the distance from the origin to the plane $z=a$ in terms of $\phi$. This is our $\rho_1$: $\rho_1=\frac{a}{\cos\phi}$
For $\rho_2$, we need to find a point on the surface of the sphere. For that, we use the equation of the sphere, which is re-written at the top left of the picture, and make our substitutions $\rho^2=x^2+y^2+z^2$ and $z=r\cos\phi$ to arrive at:
$\rho_2^2=2a\rho\cos\phi$
and thus
$\rho_2=2a\cos\phi$
We're done! The integral is now ready for you to integrate:
$ \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\pi/4}\int_{\rho=\frac{a}{\cos\phi}}^{\rho=2a\cos\phi}\rho^2\sin\phi d\rho d\phi d\theta $
I've checked this answer with Mathematica, and indeed we arrive at the expected result $\frac{2\pi a^2}{3}$.
Hope this helps. I didn't work out the integration because I assume you can do that step. It should be straightforward - you might need to look up/re-derive a couple simple trigonometric integrals.