Answer
The bat hears a sound frequency of 31.1 kHz
Work Step by Step
This is a situation with two Doppler shifts. First, we can find the frequency detected at the wall, which is a stationary object.
$f' = \frac{f}{(1-\frac{v_{source}}{v_{snd}})}$
$f' = \frac{30.0~kHz}{(1-\frac{6.0~m/s}{343~m/s}~)}$
$f' = 30.53~kHz$
Now we can find the frequency detected by the bat.
$f'' = (1+\frac{v_{obs}}{v_{snd}})~f'$
$f'' = (1+\frac{6.0~m/s}{343~m/s})~(30.53~kHz)$
$f'' = 31.1~kHz$
The bat hears a sound frequency of 31.1 kHz.
