Answer
$$\frac{\pi}{2}$$
Work Step by Step
\begin{align*}
I_{y}&=\int_{\pi}^{2 \pi} \int_{0}^{\left(\sin ^{2} x\right) / x^{2}} x^{2} d y d x\\
&=\int_{\pi}^{2 \pi}\left(\sin ^{2} x-0\right) d x\\
&=\frac{1}{2} \int_{\pi}^{2 \pi}(1-\cos 2 x) d x\\
&=\frac{1}{2} \left(x- \frac{1}{2}\sin 2x\right)\bigg|_{\pi}^{2 \pi} \\
&=\frac{\pi}{2}
\end{align*}
