The total moment of inertia of the system is the sum of the moments of inertia of the three individual spherical shells about the given axis $XX'$. The axis $XX'$ passes through the diameter of one shell and is tangent to the other two shells. Moment of inertia of a thin spherical shell about its diameter is $I_{diameter} = \frac{2}{3}mr^2$. Moment of inertia of a spherical shell about a tangent can be found using the parallel axis theorem: $I_{tangent} = I_{diameter} + mr^2 = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2$. Therefore, the total moment of inertia of the system is: $I_{total} = I_{diameter} + 2 \times I_{tangent}$ $I_{total} = \frac{2}{3}mr^2 + 2 \times \left(\frac{5}{3}mr^2\right) = \frac{2}{3}mr^2 + \frac{10}{3}mr^2 = \frac{12}{3}mr^2 = 4mr^2$