1. Use the definition of mutual inductance

Mutual inductance is

$M=\frac{\Phi}{I}$

where $\Phi$ is the magnetic flux linked with one loop due to current $I$ in the other loop.

Since $L\gg R$, the small circular loop is very small compared to the square loop, so the magnetic field due to the square loop may be taken approximately uniform over the area of the small circular loop and equal to the field at the common centre.

2. Magnetic field at the centre of the square loop

Let current $I$ flow in the large square loop of side $L$.

Distance of the centre from each side of the square is

$a=\frac{L}{2}$

For one straight side, magnetic field at the centre is given by the finite wire formula:

$B_1=\frac{\mu_0 I}{4\pi a}(\sin\theta_1+\sin\theta_2)$

Here, by symmetry,

$\theta_1=\theta_2=45^\circ$

So,

$B_1=\frac{\mu_0 I}{4\pi (L/2)}(\sin 45^\circ+\sin 45^\circ)$

=\frac\mu_0 I{2\pi L}\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)

=\frac\mu_0 I{2\pi L}(\sqrt2)

$=\frac{\sqrt2\,\mu_0 I}{2\pi L}$

Since there are 4 sides, total field at the centre is

$B=4B_1=4\cdot \frac{\sqrt2\,\mu_0 I}{2\pi L}$

$B=\frac{2\sqrt2\,\mu_0 I}{\pi L}$

3. Flux through the small circular loop

Area of the small circular loop is

$A=\pi R^2$

Hence flux through it is

$\Phi=BA=\frac{2\sqrt2\,\mu_0 I}{\pi L}\cdot \pi R^2$

$\Phi=\frac{2\sqrt2\,\mu_0 I R^2}{L}$

4. Compute mutual inductance

$M=\frac{\Phi}{I}=\frac{2\sqrt2\,\mu_0 R^2}{L}$

5. Match with the options

This corresponds to

$\boxed{M=\frac{2\sqrt2\,\mu_0 R^2}{L}}$

So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.