Answer
$14.25\space s$
Work Step by Step
Please see the attached image first.
We know that,
$Acceleration\space (a)=\frac{V^{2}}{r}\space in\space circular\space motion$
Let's plug known values into this equation.
$a=\frac{V^{2}}{r}$
$3.7\space m/s^{2}=\frac{V^{2}}{19\space m}$
$V=8.38\space m/s$
We know, the rotation period (T) of circular motion $=\frac{2\pi}{\omega}$ & $\omega = \frac{V}{r}$
So we can write,
$T=\frac{2\pi}{(\frac{V}{r})}= \frac{2\pi r}{V}$
Let's plug known values into this equation.
$T=\frac{2\times\pi \times19\space m}{8.38\space m/s}= 14.25\space s$
