A particle on a stretched string supporting a travelling wave takes...

The time taken by a particle to travel from its mean position to the extreme position is one-fourth of the time period $T$.

$\dfrac{T}{4} = 5.0 \text{ ms} \Rightarrow T = 20.0 \text{ ms} = 20 \times 10^{-3} \text{ s}$

The frequency $f$ is given by:

$f = \dfrac{1}{T} = \dfrac{1}{20 \times 10^{-3}} = 50 \text{ Hz}$

The distance separating two consecutive particles that are simultaneously at their mean positions corresponds to half of the wavelength $\lambda$.

$\dfrac{\lambda}{2} = 2.0 \text{ cm} \Rightarrow \lambda = 4.0 \text{ cm} = 0.04 \text{ m}$

The wave speed $v$ is the product of frequency and wavelength:

$v = f \lambda = 50 \times 0.04 = 2.0 \text{ m s}^{-1}$