Here, $M = 20 \,kg$
$R = 20 \,cm = 20 \times 10^{-2}\, m, \omega = 100\, rad\, s^{-1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{MR^2} {2} = \frac{(20\,kg)(20\times 10^{-2} m)^2}{2} = 0.4\,kg\,m^2$
Angular momentum of the cylinder about its axis is
$L = I\omega = (0.4\, kg \,m^2) (100\, rad\, s^{ -1}) = 40 \,J \,s$