S.O.S. Mathematics CyberBoard

Four small spheres, each of which can be regarded as a point of mass 0.200 kg, are arranged in a sphere 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis:
a.) through the center of the square, perpendicular to its plane (an axia through point O in the figure;
b.) bisecting two opposite sides of the square (an axis along the line AB in the figure);
c.) that passes through the centers of the upper left and lower right spheres and through point O.

For a point mass, the moment of intertia, I = m*r^2, where m is the mass of the object and r is the distance of the mass from its center of rotation.

I assume you meant that all four masses are arranged in a square.

a. Each mass is 0.4/sqrt(2) m from the square's center.
I{system} = 4*m*r^2
I = 4(0.2 kg)[0.4/sqrt(2) m]^2
I = 0.064 kg*m^2

b. Each mass is 0.2 m from the axis of rotation
I = 4(0.2 kg)(0.2 m)^2
I = 0.032 kg*m^2

c. Only two masses (upper right and lower left) will affect the rotational inertia since the axis of rotation passes through the upper left and lower right masses (their radius = 0). Those two masses are at a distance
r = 0.4/sqrt(2) m from the axis of rotation.
I = 2*(0.2 kg)[0.4/sqrt(2) m]^2 = 0.032 kg*m^2