A string of negligible thickness is wound multiple times around a...

Let the velocity of the center of mass of the cylinder be $v$ and its angular velocity be $\omega$.

For pure rolling on a stationary horizontal surface, the condition is $v = \omega R$, where $R$ is the radius of the cylinder.

The string is pulled from the topmost point of the cylinder. The velocity of this point is the vector sum of the translational velocity and the tangential velocity due to rotation:

$v_{\text{top}} = v + \omega R = v + v = 2v$

The cylinder reaches the man when its center of mass has travelled a distance $l$. The time taken is $t$, so $\int_{0}^{t} v \, dt = l$.

The length of the string that passes through the man's hands is the total distance moved by the end of the string (the topmost point of the cylinder) in the same time $t$.

Distance $= \int_{0}^{t} v_{\text{top}} \, dt = \int_{0}^{t} 2v \, dt = 2 \int_{0}^{t} v \, dt = 2l$.