Ans.

When a circular disc of mass M and radius r rolls on a table, then its total kinetic energy E will be,

$E$ = (Energy due to linear motion) + (energy due to rotational motion)

The moment of inertia of a circular disc about an axis passing through the centre of the disc and perpendicular to the plane of the disc is
$I=\frac{1}{2}Mr^2$
[ To know the derivation (CLICK HERE) ]

Energy due to the linear motion is $E_l=\frac{1}{2}Mv^2\\=\frac{1}{2}M{(r\omega)}^2\\=\frac{1}{2}Mr^2{\omega}^2$

Energy due to the rotational motion is $E_r=\frac{1}{2}I{\omega}^2$, [ To know the derivation of rotational kinetic energy, (CLICK HERE) ]

Putting the value of $I$ we get, $E_r=\frac{1}{4}Mr^2{\omega}^2$.

So the total kinetic energy of the disc is

$E$=$E_l+E_r\\=\frac{1}{2}Mr^2{\omega}^2+\frac{1}{4}Mr^2{\omega}^2\\=\frac{3}{4}Mr^2{\omega}^2$