Answer
$u=68.6m/s$
Work Step by Step
We can find the required speed of the train as follows:
The observed frequency of the train as it approaches the signal is given as
$f=(1+u/v)f_{\circ}$....eq(1)
and the observed frequency as the train leaves the signal is given as
$\frac{2}{3}f=(1-u/v)f_{\circ}$........eq(2)
Dividing equation (2) by equation (1), we obtain:
$\frac{\frac{2}{3}f}{f}=\frac{(1-u/v)f_{\circ}}{(1+u/v)f_{\circ}}$
This simplifies to:
$2(v+u)=3(v-u)$
$u=\frac{1}{5}v$
We plug in the known values to obtain:
$u=\frac{1}{5}(343)$
$u=68.6m/s$
