Answer
$$\dfrac{\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^\pi _0 \int^{(1-\cos\phi)/2}_0 \space p^2 \space \sin\phi \space dp \space d\phi \space d\theta =\dfrac{1}{24} \times \int^{2\pi}_0 \int^{\pi}_0 (1- \cos\phi)^3 sin\phi \space d\phi \space d\theta \\=\dfrac{1}{96} \times \int^{2\pi}_0 [(1-cos\phi)]^\pi_0 d\theta \\=\dfrac{16}{96} \times \int^{2\pi}_0 d\theta \\=\dfrac{\pi}{3}$$