Answer
The received sound frequency is 42.6 kHz
Work Step by Step
This is a situation with two Doppler shifts. First, we can find the frequency received by the moving object;
$f' = (1-\frac{v_{obs}}{v_{snd}})~f$
$f' = (1-\frac{27.5~m/s}{343~m/s})~(50.0~kHz)$
$f' = 45.99~kHz$
Now we can find the frequency detected by the bat;
$f'' = \frac{f'}{(1+\frac{v_{source}}{v_{snd}})}$
$f'' = \frac{45.99~kHz}{(1+\frac{27.5~m/s}{343~m/s}~)}$
$f'' = 42.6~kHz$
The received sound frequency is 42.6 kHz.