For the arrangement shown in the figure, the string has a mass of...

The total length of the string is the sum of the lengths on both sides of the pulley.

$l = 2.0 \text{ m} + 0.25 \text{ m} = 2.25 \text{ m}$

The mass of the string is given as $m = 4.5 \text{ g} = 4.5 \times 10^{-3} \text{ kg}$.

The linear mass density $\mu$ of the string is:

$\mu = \dfrac{m}{l} = \dfrac{4.5 \times 10^{-3}}{2.25} = 2 \times 10^{-3} \text{ kg/m}$

The tension $T$ in the string is provided by the weight of the suspended block. Since the mass of the string is negligible compared to the block, the tension is uniform throughout the string.

$T = Mg = 2 \times 10 = 20 \text{ N}$

The speed $v$ of the transverse wave in the string is:

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{20}{2 \times 10^{-3}}} = \sqrt{10000} = 100 \text{ m/s}$

The distance $d$ from the floor to the pulley is $2.0 \text{ m}$. The time $t$ required for the disturbance to reach the pulley is:

$t = \dfrac{d}{v} = \dfrac{2.0}{100} = 0.02 \text{ s}$