This problem corresponds directly to Example 6.8 in the NCERT textbook .

1. Setup: A current $I_1$ flows through the larger outer loop of radius $R_1$.
2. Magnetic Field: The magnetic field $B_1$ produced at the center of the large loop is given by $B_1 = \frac{\mu_0 I_1}{2R_1}$. Since $R_1 \gg R_2$, this field is assumed to be uniform over the area of the smaller inner loop.
3. Magnetic Flux: The magnetic flux $\Phi_2$ linked with the smaller loop (radius $R_2$) is the product of the field and the area of the small loop: $\Phi_2 = B_1 A_2 = \left(\frac{\mu_0 I_1}{2R_1}\right) (\pi R_2^2)$.
4. Mutual Inductance: By definition, $\Phi_2 = M I_1$. Comparing the expressions, we get $M = \frac{\mu_0 \pi R_2^2}{2R_1}$.
5. Proportionality: Therefore, $M \propto \frac{R_2^2}{R_1}$.