Answer
$I_y=I_x= \dfrac{\pi \rho a^4}{4} $
Work Step by Step
The parameterize representation for the curve is: $x= \cos \theta; y= 2 \sin \theta$ and $ 0 \leq \theta \lt 2 \pi $
Moment of inertia about $x$-axis can be found as: $I_x=\dfrac{-\rho}{3} \oint_{C} y^3 dx \\= \dfrac{-\rho}{3} \int_{0}^{2 \pi} ( a\sin^3 \theta) \times (-a \sin \theta d \theta)
\\=\dfrac{\rho^4}{12}\times \int_{0}^{2 \pi} (1-\cos \theta)^2 d \theta \\ =\dfrac{\rho^4}{24} \times [3 \theta-2 \sin 2 \theta+\dfrac{\sin 4 \theta}{4}]_0^{2 \pi} \\= \dfrac{\pi \rho a^4}{4} $
Therefore, by symmmerty we can say that the moment of inertia about $y$-axis will be: $I_y=I_x= \dfrac{\pi \rho a^4}{4} $
