Four thin rods of same mass  $M$  and same length $l,$ form a square as shown in the figure. Moment of inertia of this system about an axis through centre $O$ and perpendicular to its plane is

Moment of inertia of rod $AB$ about point $P=\frac{1}{12}M{l}^2$

M.I. of rod $AB$ about point $O=\frac{M{l}^2}{12}+M{\left(\frac{l}{2}\right)}^2=\frac{1}{3}M{l}^2$

[By the theorem of parallel axis]

And the system consists of 4 rods similar type so by the symmetry $I_{System}=\frac{4}{3}M{l}^2$