Answer
$I_x = \pi a^3$ , $I_y = \pi a^3$
Work Step by Step
The wire lies along $\textbf r (t) = a\cos t \textbf i + a\sin t \textbf j = x(t) \textbf i + y(t) \textbf j$
It follows that $x(t) a \cos t , y(t) = a\sin t$
$\textbf r' = -a\sin t \textbf i + a\cos t \textbf j \Rightarrow \vert \textbf r' \vert = a$
$\rho (x,y) = 1$
$I_x = \int_C y^2\rho(x,y) ds = \int_{t=0}^{2\pi}a^2\sin ^2 t.\vert \textbf r' \vert dt =\int_{t=0}^{2\pi}a^3\sin ^2t dt = a^3[\frac{x}{2}-\frac{\sin 2x}{4}]_0^{2\pi}=\pi a^3$
$I_y = \int_C x^2\rho(x,y) ds = \int_{t=0}^{2\pi}a^2\cos ^2 t\vert \textbf r' \vert dt=\int_{t=0}^{2\pi}a^3\cos ^2t dt = a^3[\frac{x}{2}+\frac{\sin 2x}{4}]_0^{2\pi} = \pi a^3$
