homework and exercises - Is the solution to a question regarding four smaller masses on a rotating hollow sphere accurate?

I was doing the question here:

A uniform 8.40-kg spherical shell, 50.0 cm in diameter has four small 2.00-kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the center of the sphere and two of the small masses (see figure). What friction torque is needed to reduce its angular speed from 75.0 rpm to 50.0 rpm in 30.0 s?

I understand how to get $\alpha$, and how to get $\tau$ from $I$ and $\alpha$. What I do not understand is their computation of the moment of inertia. The question did not provide any information regarding the shapes of the masses, but the solution (which writes $I=\frac{2}{3}MR^2+2mR^2$) seems to assume that the masses are point particles. Thus, the two masses along the rotational axis have a moment of inertia of 0 ($r=0$), and each of the other two masses has a moment of inertia of $I=mr^2 =mR^2$, where $R$ is the radius of the sphere and $r$ is the distance from the axis.

But why are the masses considered to be point particles? I wanted to use the parallel-axis theorem to determine the moment of inertia about the rotational axis; not knowing the shape of the masses made this impossible, since I couldn’t find the moment of inertia about the CM. If I knew $I_{CM}$, then couldn’t I use $I_{axis}=I_{CM}+mR^2$ for the two masses not on the rotational axis and $I_{CM}$ for the two that are? That would make $I_{total}=\frac{2}{3}MR^2+4I_{CM,\:mass}+2mR^2$ Would that not be a better way of solving this problem?