Answer
$$\dfrac{3}{8}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Average= \dfrac{1}{(2 \pi/3)} \times \int^{2\pi}_0 \int^{\pi/2}_0 \int^1_0 p^3 \cos(\phi) \space \sin(\phi) \space dp \space d\phi \space d\theta \\=\dfrac{3}{8\pi} \times \int^{2\pi}_0 \int^{\pi/2}_0 \cos(\phi) \sin( \phi) \space d\phi \space d\theta \\=\dfrac{3}{8\pi} \times \int^{2\pi}_0 [\dfrac{sin^2\phi}{2}]^{\pi/2}_0 (1) d\theta \\=\dfrac{3}{16\pi} \times \int^{2\pi}_0 (1) d\theta \\=(\dfrac{3}{16\pi})(2\pi)\\=\dfrac{3}{8}$$
