A very small conducting circular loop of radius r is placed in the...

Let the current in the long straight wire be $I$.

The magnetic field produced by the long straight wire at a perpendicular distance $x$ is given by $B = \dfrac{\mu_0 I}{2 \pi x}$.

Since $x \gg r$, the magnetic field over the area of the small circular loop can be considered approximately uniform and equal to the value at its center.

The area of the circular loop is $A = \pi r^2$.

The magnetic flux $\Phi$ passing through the loop is $\Phi = B A = \left( \dfrac{\mu_0 I}{2 \pi x} \right) (\pi r^2) = \dfrac{\mu_0 I r^2}{2 x}$.

The mutual inductance $M$ is defined as the magnetic flux through the loop per unit current in the wire, so $M = \dfrac{\Phi}{I} = \dfrac{\mu_0 r^2}{2 x}$.

Therefore, the mutual inductance $M$ is directly proportional to $\dfrac{r^2}{x}$.

Answer: $\dfrac{r^2}{x}$