Goal

Mutual inductance is $M=\frac{\Phi}{I}$ where $\Phi$ is the magnetic flux linked with the small loop due to current $I$ in the large loop.

Use the fact that $b\gg a$

Since the small loop is very small compared to the large one and both are concentric, the magnetic field over the small loop may be taken approximately uniform and equal to the field at the center of the large square loop.

So, $\Phi = B_{\text{center}}\cdot a^2$ because the area of the small square loop is $a^2$.

Magnetic field at the center of the large square loop

The large square has side $b$. Distance of its center from each side is $r=\frac b2$

Magnetic field due to one finite straight side at a point on its perpendicular bisector is $B_{1}=\frac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2)$

Here, by symmetry, $\theta_1=\theta_2=45^\circ$ since half the side is $b/2$ and the perpendicular distance is also $b/2$.

Therefore for one side, $B_1=\frac{\mu_0 I}{4\pi (b/2)}\left(\sin45^\circ+\sin45^\circ\right)$ $=\frac{\mu_0 I}{4\pi (b/2)}\left(2\cdot \frac{1}{\sqrt2}\right)$ $=\frac{\mu_0 I}{4\pi (b/2)}\sqrt2$ $=\frac{\mu_0 I}{4\pi}\frac{2\sqrt2}{b}$

Add contributions of all four sides

All four sides produce magnetic field in the same perpendicular direction at the center, so $B_{\text{center}}=4B_1$ $B_{\text{center}}=4\cdot \frac{\mu_0 I}{4\pi}\frac{2\sqrt2}{b}$ $B_{\text{center}}=\frac{\mu_0 I}{4\pi}\frac{8\sqrt2}{b}$

Flux through the small loop

Since the field is approximately uniform over the small loop, $\Phi = B_{\text{center}}\, a^2$ $\Phi = \left(\frac{\mu_0 I}{4\pi}\frac{8\sqrt2}{b}\right)a^2$

Mutual inductance

$M=\frac{\Phi}{I}$ $M=\frac{1}{I}\left(\frac{\mu_0 I}{4\pi}\frac{8\sqrt2 a^2}{b}\right)$ $M=\frac{\mu_0}{4\pi}8\sqrt2\frac{a^2}{b}$

Match with options

This corresponds to: $\boxed{\frac{\mu_0}{4\pi}8\sqrt2\frac{a^2}{b}}$

So the correct option is A.