Answer
$$\dfrac{2}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Average=\dfrac{1}{2\pi} \int^{2\pi}_0 \int^1_0 \int^1_{-1} 2r^2 \times dr \space d\theta \\=\dfrac{1}{2\pi}\int^{2\pi}_0 \int^1_0 (2r^2) \space dr \space d \theta \\=\dfrac{1}{3\pi} \times \int^{2\pi}_0 d\theta \\=\dfrac{2}{3}$$
