Answer
$$1.022 \times 10^{3} \mathrm{Hz}$$
Work Step by Step
We mark speed of French submarine by $u_{1}$
and U.S. sub by $u_{2}$ .
Therefore frequency, as detected by the U.S. sub, is
$$f_{1}^{\prime}=f_{1}\left(\frac{v+u_{2}}{v-u_{1}}\right)$$$$=\left(1.000 \times 10^{3} \mathrm{Hz}\right)\left(\frac{5470 \mathrm{km} / \mathrm{h}+70.00 \mathrm{km} / \mathrm{h}}{5470 \mathrm{km} / \mathrm{h}-50.00 \mathrm{km} / \mathrm{h}}\right)=1.022 \times 10^{3} \mathrm{Hz}$$
