Let us consider one of the four rods. The moment of inertia of a thin rod of mass $M$ and length $t$ about an axis passing through its own centre of mass and perpendicular to its length is $I_{CM} = \frac{Mt^2}{12}$. The perpendicular distance from the centre of the square to the centre of each rod is $d = \frac{t}{2}$. According to the parallel axis theorem, the moment of inertia of one rod about the axis passing through the centre of the square and perpendicular to its plane is: $I_1 = I_{CM} + Md^2 = \frac{Mt^2}{12} + M\left(\frac{t}{2}\right)^2 = \frac{Mt^2}{12} + \frac{Mt^2}{4} = \frac{4Mt^2}{12} = \frac{Mt^2}{3}$ Since the square frame consists of 4 identical rods, the total moment of inertia of the frame is: $I_{total} = 4 \times I_1 = 4 \times \frac{Mt^2}{3} = \frac{4}{3}Mt^2$