Consider the square made of 4 thin uniform rods of length a.

The blue small circles in the middle of each rod represents its center of mass.

If we considere a perpendicular axis of rotation through each center of mass: \(\displaystyle{I}_{{{c}{m}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}\)

To find the moment of inertia of each rod about an axis through the center of the square and perpendicular to its plane (point P in the figure), use the Parallel - Axis Theorem: \(\displaystyle{I}_{{{p}}}={I}_{{{c}{m}}}+{\left({\frac{{{M}}}{{{4}}}}\right)}{\left({\frac{{{a}}}{{{2}}}}\right)}^{{{2}}}={\frac{{{1}}}{{{12}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}+{\frac{{{1}}}{{{4}}}}{\left({\frac{{{M}}}{{{4}}}}\right)}{a}^{{{2}}}={\frac{{{1}}}{{{12}}}}{M}{a}^{{{2}}}\)

The total moment of inertia is equal to four times the value f or each rod (the figure is symmetric): \(\displaystyle{I}_{{{p}\ to{t}{a}{l}}}={4}{I}_{{{p}}}={4}{\left({\frac{{{1}}}{{{12}}}}\right)}{M}{a}^{{{2}}}={\frac{{{1}}}{{{3}}}}{M}{a}^{{{2}}}\)