Moment of inertia of complete disc about point 'O'.

${I_{Total\,disc}} = {{M{R^2}} \over 2}$

Mass of removed disc

${M_{{\mathop{\rm Removed}\nolimits} }} = {M \over 4}\left( {Mass \propto area} \right)$

Moment of inertia of removed disc about point 'O'.

I_Removed (about same perpendicular axis)

$= {I_{cm}} + m{x^2}$

$= {M \over 4}{{{{\left( {R/2} \right)}^2}} \over 2} + {M \over 4}{\left( {{R \over 2}} \right)^2} = {{3M{R^2}} \over {32}}$

Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,

I_{Remaing disc} = I_Total – I_Removed

$= {{M{R^2}} \over 2} - {3 \over {32}}M{R^2} = {{13} \over {32}}M{R^2}$