Answer
$$1.045 \times 10^{3} \mathrm{Hz} $$
Work Step by Step
If French sub is stationary, then frequency of reflected wave would be
$$
f_{r}=f_{1}\left(v+u_{2}\right) /\left(v-u_{2}\right)
$$
since French sub moving toward reflected signal with a speed $u_{1},$ therefore
$$ f_{r}^{\prime} =f_{r}\left(\frac{v+u_{1}}{v}\right)$$$$=f_{1} \frac{\left(v+u_{1}\right)\left(v+u_{2}\right)}{v\left(v-u_{2}\right)}$$$$=\frac{\left(1.000 \times 10^{3} \mathrm{Hz}\right)(5470+50.00)(5470+70.00)}{(5470)(5470-70.00)} =1.045 \times 10^{3} \mathrm{Hz} $$
