Answer
$$\dfrac{2a}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Average=\dfrac{4}{\pi a^2} \int^{\pi/2}_0 \int^{a}_0 r\sqrt{a^2-r^2} \space dr \space d\theta $
or, $=\dfrac{4}{3\pi a^2} \times \space \int^{\pi/2}_0 \space a^3 \space d\theta $
or, $=\dfrac{2a}{3}$