Answer
55.3 kHz.
Work Step by Step
Use equation 12–4, because the source and observer are both in motion.
There are 2 Doppler shifts in this problem: the first is for the emitted sound with the bat as the source and the moth as the observer
$$f’_{moth}=f_{bat}\frac{v_{sound}+v_{moth}}{ v_{sound}-v_{bat}}$$
The second Doppler shift is for the reflected sound, with the moth as the source and the bat as the observer.
$$f’’_{bat}=f’_{moth}\frac{v_{sound}+v_{bat}}{ v_{sound}-v_{moth}}$$
$$= f_{bat}\frac{v_{sound}+v_{moth}}{ v_{sound}-v_{bat}}\frac{v_{sound}+v_{bat}}{ v_{sound}-v_{moth}}$$
$$ = (51.35\;kHz)\frac{343+5.0}{343-7.8}\frac{343+7.8}{ 343-5.0}=55.3\;kHz$$