Answer
$$\dfrac{31}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ Average=\dfrac{1}{8} \times \int^2_0 \int^2_0 \int^2_0 (x^2+p) \space dz \space dy \space dx \\=\dfrac{1}{8} \times \int^2_0 \int^2_0 (2x^2+18) \space dy \space dx \\=\dfrac{1}{8} \times \int^2_0 (4x^2+36) \space dx \\=\dfrac{31}{3}$$
