Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by light rods

The question is missing some parts but nevertheless, the question is just asking for the moment of inertia of the system about an axis through the center of the square.

The solution would be:

I = 4*(M*R^2)

where R = 0.200m*sqrt(2) = 0.2828m

Take note use 0.200m as 1/2 the side of the square.

So therefore, I = 4 * (0.200kg*0.2828^2)

= 6.40x10^-2 kg-m^2

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explanation: