A square wire frame  ABCD has mass m and length of  each side is l. Then moment of inertia of the frame about side AB is 

 

    mass of each rod is  m/4. Moment of inertia of  AD and  BC about AB 

$2\left[\frac{m}{4}\frac{l^2}{3}{\sin }^2\left({60}^0\right)\right]$  and moment of inertia of the rod CD about AB is $\frac{m}{4}{h}^2 \left( h=l\sin {60}^0\right)$

so total moment of inertia $2\left[\frac{m}{4}\frac{l^2}{3}{\sin }^2{60}^0\right]+\frac{m}{4}{\left(l\sin {60}^0\right)}^2=\frac{m}{4}{l}^2{\sin }^{2}60\left(\frac{2}{3}+1\right)=\frac{5m{l}^2}{16}$