Explanation :
Moment of internal of cross section about horizontal centroid axis (x'-x')
$ \begin{array}{l}{\mathrm I}_\mathrm{Critical}=\frac{\mathrm\pi}{64}\mathrm d^4=\frac{\mathrm\pi}{64}\times\left(8\right)^4=201.06\;\mathrm{mm}^4\\{\mathrm I}_\mathrm{Rectangle}=\frac{\mathrm{bh}^3}{12}=\frac{10\times\left(20\right)^3}{12}=6666.67\mathrm{mm}^4\\{\mathrm I}_\mathrm{total}={\mathrm I}_\mathrm R-{\mathrm I}_\mathrm C=6666.67-201.06=6465.6\;\mathrm{mm}^4\end{array} $
M.D.I about base axis(x-x)
Applying parallel axis theorem
$ \begin{array}{l}{\mathrm I}_\mathrm{xx}={\mathrm I}_{\mathrm x'\mathrm x'}+\mathrm A\times\mathrm y^2\\{\mathrm I}_\mathrm{xx}=6465.6+\left[\left(10\times20\right)-\frac{\mathrm\pi}4\times\left(8\right)^2\right]\times\left(10\right)^2\\{\mathrm I}_\mathrm{xx}=21439.058\;\mathrm{mm}^4\end{array} $