Answer
$$\dfrac{17\pi}{5}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^{\theta/(2\pi)}_0 \int^{3+24r^3}_0 dz \space r \space d \theta = \int^{2\pi}_0 \int^{\theta/(2\pi)}(3r+24r^3) \space dr \space d\theta \\=\int^{2\pi}_0 [\dfrac{3r^2}{2}+6r^4] d\theta \\=\dfrac{3}{2}\int^{2\pi}_0 (\dfrac{\theta^2} {4\pi^2}+\dfrac{4\theta^4}{16\pi^4})^{2\pi}_0 \\=\dfrac{17\pi}{5}$$
