Explanation

The square loop consists of four identical solid cylinders.

Total mass of the loop is $M$, so mass of each cylinder is

$$ m = \frac{M}{4} $$

The axis passes through the midpoints of two opposite sides and lies in the plane of the square.

Two cylinders lie symmetrically on the axis. For a solid cylinder, moment of inertia about an axis through its centre and perpendicular to its length is

$$ I = \frac{1}{4}mR^2 + \frac{1}{12}mL^2 $$

Contribution of these two cylinders:

$$ I_1 = 2\left(\frac{1}{4}\cdot\frac{M}{4}R^2 + \frac{1}{12}\cdot\frac{M}{4}L^2\right) $$

$$ I_1 = \frac{M}{8}R^2 + \frac{M}{24}L^2 $$

The remaining two cylinders are parallel to the axis and each is at a distance $\dfrac{L}{2}$ from it.

Using parallel axis theorem, moment of inertia of one such cylinder is

$$ I = \frac{1}{4}mR^2 + m\left(\frac{L}{2}\right)^2 $$

Contribution of these two cylinders:

$$ I_2 = 2\left(\frac{1}{4}\cdot\frac{M}{4}R^2 + \frac{M}{4}\cdot\frac{L^2}{4}\right) $$

$$ I_2 = \frac{M}{8}R^2 + \frac{M}{8}L^2 $$

Total moment of inertia of the square loop:

$$ I = I_1 + I_2 $$

$$ I = \left(\frac{M}{8}R^2 + \frac{M}{24}L^2\right) + \left(\frac{M}{8}R^2 + \frac{M}{8}L^2\right) $$

$$ I = \frac{3}{8}MR^2 + \frac{1}{6}ML^2 $$

Hence, the correct answer is

$$ \boxed{\frac{3}{8}MR^2 + \frac{1}{6}ML^2} $$