In this case the observer will notice two frequencies. One, direct through the source and second, through the reflection by moving wall.

As both source and receiver are at rest the frequency of sound waves which directly reach to receiver, will be 1700 Hz.

Hence ${\mathrm{n}}_{\text{direct }}=1700\mathrm{Hz}$. The frequency of sound which wall will receive as a moving observer is

${\mathrm{n}}_1={\mathrm{n}}_0\left[\frac{\mathrm{v}-\mathrm{u}}{\mathrm{v}-0}\right]={\mathrm{n}}_0\left[\frac{\mathrm{v}-\mathrm{u}}{\mathrm{v}}\right]$                    …(i)

The wall now behaves as a moving source of frequency n₁ which when received by receiver, the frequency observed is 

${\mathrm{n}}_2={\mathrm{n}}_1\left[\frac{\mathrm{v}-0}{\mathrm{v}-(-\mathrm{u})}\right]={\mathrm{n}}_1\left[\frac{\mathrm{v}}{\mathrm{v}+\mathrm{u}}\right]={\mathrm{n}}_0\left(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{v}+\mathrm{u}}\right)$                …(ii)

$=1700\left[\frac{340-0.06}{340+0.06}\right]=1700\times \frac{339.94}{340.06}=1699.4\mathrm{Hz}$

Thus, beat frequency received by detector is 

$\mathrm{\Delta n}=1700-1699.4=0.6\mathrm{Hz}$