In the given figure , two wheels $P$ and $q$ are connected by a belt $B$. The radius of $P$ is three times as that of $Q$. In case of same rotational kinetic energy, the ratio of rotational inertias $\left(\frac{I_P}{I_Q}\right)$ will be $x:1$. The value of $x$ will be

Let ${\omega }_P$ and ${\omega }_Q$ be the angular velocity of wheels $P$ and $Q$ respectively, and since, rotational kinetic energy is same for them, the tangential velocity will also be same.
$\Rightarrow {\omega }_P\left(3R\right)={\omega }_Q\left(R\right)\Rightarrow \frac{{\omega }_Q}{{\omega }_P}=\frac{3}{1}..\left(1\right)$
Also, from the relation of rotational kinetic energy,
$\frac{1}{2}{I}_P{\left({\omega }_P\right)}^2=\frac{1}{2}{I}_Q{\left({\omega }_Q\right)}^2$
$\frac{I_P}{I_Q}={\left(\frac{{\omega }_Q}{{\omega }_P}\right)}^2=\frac{9}{1}$ [from eq $\left(1\right)$]
Hence, value of $x$ will be $9$.