$(a)$ We use $I \alpha=\tau$
the torque $\quad \tau=F R$ $=25 \times 0.20 Nm \text { (as } R=0.20 m )$
$=5.0 Nm$
$I=$ Moment of inertia of flywheel about its $\operatorname{axis}=\frac{M R^{2}}{2}$
$=\frac{20.0 \times(0.2)^{2}}{2}=0.4 kg m ^{2}$
$\alpha=$ angular acceleration $=5.0 N m / 0.4 kg m ^{2}=12.5 s ^{-2}$
$(b)$ Work done by the pull unwinding $2 m$ of the cord $=25 N \times 2 m =50 J$
$(c)$ Let $\omega$ be the final angular velocity. The kinetic energy gained $=\frac{1}{2} I \omega^{2}$
since the wheel starts from rest. Now, $\omega^{2}=\omega_{0}^{2}+2 \alpha \theta, \quad \omega_{0}=0$
The angular displacement $\theta=$ length of unwound string / radius of wheel $=2 m / 0.2 m =10 rad$
$\omega^{2}=2 \times 12.5 \times 10.0=250( rad / s )^{2}$
$\therefore \text { K.E. gained }=\frac{1}{2} \times 0.4 \times 250=50 J$
$(d)$ The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due to friction.