Answer
$$\dfrac{3\pi}{16}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Average=\dfrac{1}{(4 \pi/3)}\int^{2\pi}_0 \int^1_0 \int^{\sqrt{1-r^2}}_{-\sqrt{1-r^2}}r^2 \space dz \space dr \space d \theta \\=\dfrac{3}{4\pi} \times \int^{2\pi}_0 \int^1_0 2r^2 \sqrt{1-r^2} \space dr \space d\theta \\=\dfrac{3}{16\pi} \times \int^{2\pi}_0 (\dfrac{\pi}{2}+0) \space d\theta \\=\dfrac{3\pi}{16}$$
