One method, bringing you to an approximate frequency directly:
The path difference of the two waves is 1.2 - 0.8 = 0.4 m
As jcaron said, you get a minima when the path difference is half of a wavelength, 1.5 of a wavelength, etc, that is when you have (0.5 + n) of a wavelength as path difference, that is
Now, this equals 0.4.
Then, recall the formula
Take f = 20000 Hz
Substitute lambda from what you got earlier to get:
Solve for n, the number of wavelengths to get an idea of how many n there are.
You'll get n = 23.029
Now, use n = 23 (n needs to be an integer) [You might also want to try one unit above and below to confirm, that is, try n = 22 and n = 24]