A string of length 20 cm and linear mass density 0.40 g cm^ -1 is...

Given, length of the string, $L = 20 \text{ cm} = 0.2 \text{ m}$
Linear mass density, $\mu = 0.40 \text{ g cm}^{-1} = \dfrac{0.40 \times 10^{-3} \text{ kg}}{10^{-2} \text{ m}} = 0.04 \text{ kg m}^{-1}$
Tension in the string, $T = 16 \text{ N}$

The speed of the transverse wave pulse on the string is given by:
$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{16}{0.04}} = \sqrt{400} = 20 \text{ m s}^{-1}$

For the string to again exhibit the exact same shape (an upright pulse traveling towards the right), the pulse must undergo two reflections.
1. It travels to the right end, reflects, and becomes inverted (since the end is fixed).
2. It travels back to the left end, reflects again, and becomes upright, traveling to the right.

The total distance traveled by the pulse to return to its initial state is $2L$.
Distance $d = 2 \times 0.2 \text{ m} = 0.4 \text{ m}$

The time taken is:
$t = \dfrac{d}{v} = \dfrac{0.4}{20} = 0.02 \text{ s}$