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(C) The question asks for the moment of inertia of the system of four spheres about axes passing through the centers $A, B, C$, and $D$.Let $I_{cm} =

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$I_{A} = I_{D} > I_{B} = I_{C}$

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(C) The question asks for the moment of inertia of the system of four spheres about axes passing through the centers $A, B, C$, and $D$.Let $I_{cm} = \frac{2}{5}MR^2$ be the moment of inertia of a single sphere about its own center.Using the parallel axis theorem, the moment of inertia of the system about an axis passing through $A$ is:$I_A = I_{cm} + (I_{cm} + M(2R)^2) + (I_{cm} + M(4R)^2) + (I_{cm} + M(6R)^2) = 4I_{cm} + 56MR^2$Similarly, about an axis passing through $B$:$I_B = (I_{cm} + M(2R)^2) + I_{cm} + (I_{cm} + M(2R)^2) + (I_{cm} + M(4R)^2) = 4I_{cm} + 24MR^2$About an axis passing through $C$:$I_C = (I_{cm} + M(4R)^2) + (I_{cm} + M(2R)^2) + I_{cm} + (I_{cm} + M(2R)^2) = 4I_{cm} + 24MR^2$About an axis passing through $D$:$I_D = (I_{cm} + M(6R)^2) + (I_{cm} + M(4R)^2) + (I_{cm} + M(2R)^2) + I_{cm} = 4I_{cm} + 56MR^2$From the above calculations, we get $I_A = I_D$ and $I_B = I_C$.Also, $56MR^2 > 24MR^2$, so $I_A = I_D > I_B = I_C$.

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