A string with a total length of 30 cm and a linear mass density of...

Given the wave speed $v = 20 \text{ cm s}^{-1}$ and the initial distance of the pulse peak from the ring is $d = 20 \text{ cm}$.

The time taken by the pulse to reach the ring is $t_1 = \dfrac{d}{v} = \dfrac{20}{20} = 1 \text{ s}$.

Since the ring is free to slide, it acts as a free boundary. Reflection at a free boundary occurs without any phase reversal, so the upright pulse reflects as an upright pulse moving in the opposite direction.

For the string to regain its initial spatial shape, the symmetric upright pulse must return to its original position.

The distance traveled back to the initial position is again $d = 20 \text{ cm}$.

The time taken for this return journey is $t_2 = \dfrac{d}{v} = \dfrac{20}{20} = 1 \text{ s}$.

The total time taken is $t = t_1 + t_2 = 1 + 1 = 2 \text{ s}$.