1. Given

   • Two wheels $P$ and $Q$ are connected by a belt, so the linear speed at the rim is the same: $v_P=v_Q$
   • Radius of $P$ is three times radius of $Q$: $R_P=3R_Q$
   • Rotational kinetic energies are equal: $\frac12 I_1\omega_1^2=\frac12 I_2\omega_2^2$

2. Use belt condition Since rim speeds are equal, $\omega_1 R_P=\omega_2 R_Q$ Substituting $R_P=3R_Q$: $\omega_1(3R_Q)=\omega_2 R_Q$ $3\omega_1=\omega_2$ $\omega_2=3\omega_1$

3. Use equality of rotational kinetic energies $I_1\omega_1^2=I_2\omega_2^2$ Substitute $\omega_2=3\omega_1$: $I_1\omega_1^2=I_2(3\omega_1)^2$ $I_1\omega_1^2=9I_2\omega_1^2$ $I_1=9I_2$

4. Find the ratio $\frac{I_1}{I_2}=9$ So the ratio is $I_1:I_2=9:1$

Therefore, the required value is: $\boxed{9}$