A cylindrical drum, pushed along by a board rolls forward on the ground. There is no slipping at any contact. The distance moved by the man who is pushing the board, when axis of the cylinder covers a distance L will be

Let ${\mathrm{v}}_0$ be the linear speed of the axis of the cylinder and $\mathrm{\omega }$ be its angular speed about the axis. As it does not slip on the ground hence $\text{ω = }\frac{{\mathrm{v}}_0}{\mathrm{R}}$. Where R is the radius of the cylinder.  

Speed of the topmost point is v = ${\mathrm{v}}_0+\mathrm{\omega R} = 2{\mathrm{v}}_0$
Since time taken by the axis to move a distance L is equal to t = $\frac{\mathrm{L}}{{\mathrm{v}}_0}$.
In the same interval of time distance moved by the topmost point is $\mathrm{S} = 2{\mathrm{v}}_0\times \frac{\mathrm{L}}{{\mathrm{v}}_0} = 2\mathrm{L}$