The moment of inertia of a system of four rods, each of length I and mass m, about the axis perpendicular to plane of square as shown is 

Moment of inertia about axis 2 which is parallel to axis 1:

$I_2=4\left(\frac{m{l}^2}{12}+m{\left(\frac{l}{2}\right)}^2\right)=\frac{4m{l}^2}{3}$

Apply perpendicular axis theorem,

$\begin{array}{c}{I}_1=I_2+4m{x}^2\\ =\frac{4m{l}^2}{3}+4m{\left(\frac{l}{\sqrt{2}}\right)}^2=\frac{10}{3}m{l}^2\end{array}$