The moment of inertia of a hollow cubical box of mass M and side a about an axis passing through the centres of two opposite faces is equal to

Taking mass of plate m

       Then MI of two plates through which axis is passing  = $\frac{\mathrm{m} {\mathrm{a}}^2}{6}\times 2 = \frac{{\mathrm{ma}}^2}{3}$

M.I. of 4 plates having symmetrical position from the axis.

= $4\times [\frac{{\mathrm{ma}}^2}{12}+\mathrm{m}{\left(\frac{\mathrm{a}}{2}\right)}^2] = 4\times \left[\frac{{\mathrm{ma}}^2}{3}\right]$

Total MI = $\frac{4 {\mathrm{ma}}^2}{3}+\frac{{\mathrm{ma}}^2}{3} = \frac{5{\mathrm{ma}}^2}{3}$

using $\mathrm{m} =\frac{M}{6} \Rightarrow \mathrm{MI} = \frac{5 {\mathrm{Ma}}^2}{18}$