Solution:

Given,
Velocity of an observer (uo) = 20 m/s
Velocity of source of sound (us) = 0
Change of frequency = 50 Hz
Speed of sound in the air (v) = 340 m/s
Frequency of teh source (f) = ?

We know, from the doppler's effect formula that,

\(f_1'=\left(\frac{v\pm u_0}{v\pm u_s}\right)\times f\)

In this condition, when the observer approaches teh source, we have

\(f_1'=\left(\frac{v+v_o}v\right)\times f=\left(\frac{340+20}{240}\right)\times f=\frac{18}{17}f\)

Again, when an observer passes the source,we have

\(f_2'=\left(\frac{v-v_o}v\right)\times f=\left(\frac{340-20}{240}\right)\times f=\frac{16}{17}f\)

Then according to the question, 

\(f_1'-f_2'=\frac{18}{17}f-\frac{16}{17}f=50\)
Or, \(50=\frac{18f-16f}{17}\)
Or, \(\frac{2f}{17}=50\)
∴ f = 425 Hz

Hence, the required frequency is 425 Hz.