Answer
The length of the string is 2.0 meters.
Work Step by Step
We can find the length of the string as:
$v = \sqrt{\frac{F_T}{\mu}}$
$\frac{L}{t} = \sqrt{\frac{F_T}{(m/L)}}$
$\frac{L}{t} = \sqrt{\frac{F_T~L}{m}}$
$\frac{L^2}{t^2} = \frac{F_T~L}{m}$
$L = \frac{F_T~t^2}{m}$
$L = \frac{(20~N)(50\times 10^{-3}~s)^2}{0.025~kg}$
$L = 2.0~m$
The length of the string is thus 2.0 meters.
