For disc, $I=\frac{1}{2} m a^{2}$
For ring, $I=m a^{2}$
For square of side $2 a$
$=\frac{M}{12}\left[(2 a)^{2}+(2 a)^{2}\right]=\frac{2}{3} M a^{2}$
For square of rod of length $2 a$
$I=4\left[M \frac{(2 a)^{2}}{12}+M a^{2}\right]=\frac{16}{3} M a^{2}$
Hence, moment of inertia is maximum for square of four rods.