To solve this problem, we need to find the moment of inertia of the system about the $XX'$ axis.$\\$Given:$\\$• Three identical spherical shells, each of mass $m$ and radius $r. \\$The moment of inertia of a single spherical shell about an axis through its center is:$\\ I_{center} = \frac{2}{3} m r^2 \\$Let's analyze the configuration:$\\$• The axis $XX'$ passes through the diameter of the top shell and touches the other two shells.$\\$1. Moment of inertia of the top shell:$\\$Since the axis passes through its diameter, the moment of inertia is $I_{center}. \\$2. Moment of inertia of the bottom two shells:$\\$• The axis is at a distance $r$ from the center of each of these shells.$\\$• Use the parallel axis theorem:$\\ I_{parallel} = I_{center} + m d^2 \\$where $d = r. \\$For each bottom shell:$\\ I_{bottom} = \frac{2}{3} m r^2 + m r^2 = \frac{5}{3} m r^2 \\$3. Total moment of inertia of the system:$\\ I_{total} = I_{center} + 2 \times I_{bottom} \\ I_{total} = \frac{2}{3} m r^2 + 2 \times \frac{5}{3} m r^2 \\ I_{total} = \frac{2}{3} m r^2 + \frac{10}{3} m r^2 \\ I_{total} = \frac{12}{3} m r^2 \\ I_{total} = 4 m r^2 \\$Therefore, the moment of inertia of the system about the $XX'$ axis is $4 m r^2. \\$This corresponds to Option 3.