Answer
We can photograph the wave in position A at $7.2~ms$ and $14.4~ms$
We can photograph the wave in position B at $1.8~ms$ and $5.4~ms$
We can photograph the wave in position C at $3.6~ms$ and $10.8~ms$
Work Step by Step
We can find the wave speed along the string:
$v = \sqrt{\frac{F}{\mu}}$
$v = \sqrt{\frac{2.00~N}{0.200\times 10^{-3}~kg/m}}$
$v = 100~m/s$
From position A to position B, the wave peak travels $\frac{1}{4}$ of the length of the string. We can find the time for the wave to travel $\frac{1}{4}$ of the length of the string:
$t = \frac{L/4}{v}$
$t = \frac{L}{4v}$
$t = \frac{0.720~m}{(4)(100~m/s)}$
$t = 0.0018~s$
$t = 1.8~ms$
After another $1.8~ms$, the wave is in position C. We can photograph the wave in position C at $3.6~ms$
After another $1.8~ms$, the wave is in position B. We can photograph the wave in position B at $5.4~ms$
After another $1.8~ms$, the wave is in position A. We can photograph the wave in position A at $7.2~ms$
After another $1.8~ms$, the wave is in position B. We can photograph the wave in position B at $9.0~ms$
After another $1.8~ms$, the wave is in position C. We can photograph the wave in position C at $10.8~ms$
After another $1.8~ms$, the wave is in position B. We can photograph the wave in position B at $12.6~ms$
After another $1.8~ms$, the wave is in position A. We can photograph the wave in position A at $14.4~ms$
We can photograph the wave in position A at $7.2~ms$ and $14.4~ms$
We can photograph the wave in position B at $1.8~ms$ and $5.4~ms$
We can photograph the wave in position C at $3.6~ms$ and $10.8~ms$