Four identical thin rods each of mass $M$ and length $l$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is:

1. Consider one rod of the frame. Its moment of inertia about its own center of mass (perpendicular to its length) is $I_{cm}=\frac{M{l}^2}{12}$.2. The distance from the center of the square to the center of each rod is $r=\frac{l}{2}$.3. Using the Parallel Axis Theorem, the moment of inertia of one rod about the center of the square is:
$$I_{rod}=I_{cm}+M{r}^2=\frac{M{l}^2}{12}+M{\left(\frac{l}{2}\right)}^2$$
$$I_{rod}=\frac{M{l}^2}{12}+\frac{M{l}^2}{4}=\frac{M{l}^2+3M{l}^2}{12}=\frac{4M{l}^2}{12}=\frac{1}{3}M{l}^2$$
4. Since there are four such rods, the total moment of inertia is:
$$I_{total}=4\times I_{rod}=4\times \frac{1}{3}M{l}^2=\frac{4}{3}M{l}^2$$