Answer
$$\dfrac{7\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume=4\int_{0}^{\pi/2} \int^{\pi/4}_0 \int^{2\sec\phi}_{\sec\phi}p^2 \space \sin\phi dp \space d\phi \space d\theta \\=\dfrac{4}{3}\int_{0}^{\pi/2} \int^{\pi/4}_0 (8sec^3\phi-\sec^3\phi)\sin\phi \space d\phi \space d\theta \\=\dfrac{28}{3} \times \int_{0}^{\pi/2} \int^{\pi/4}_0 \tan\phi \sec^2\phi d\phi d\theta \\=\dfrac{28}{3} \times \int_{0}^{\pi/2} [\frac{1}{2}tan^2\phi]^{\pi/4}_0 d\theta \\=\dfrac{14}{3} \times \int_{0}^{\pi/2} (1) d\theta \\=\dfrac{7\pi}{3}$$
