A siren on a pole by a straight road sounds a steady note of 510 Hz…

Step 1. Speed (a). Moving away from the siren the cyclist hears $f\dfrac{c - u}{c}$; moving towards the wall, $f\dfrac{c + u}{c}$. The beat frequency is $\dfrac{2fu}{c}$, so

Step 2. The two notes (b, c). $f_d = 510\ \text{Hz} \times \dfrac{338\ \text{m/s}}{340\ \text{m/s}} = 507.0\ \text{Hz}$ and $f_e = 510\ \text{Hz} \times \dfrac{342\ \text{m/s}}{340\ \text{m/s}} = 513.0\ \text{Hz}$.

Step 3. The echo in the wind (d). Between two bodies at rest a wind does not change the frequency, so the wall still sends back $510\ \text{Hz}$. The echo travels against the wind at $330\ \text{m/s}$ over the ground, and the cyclist meets it at $332\ \text{m/s}$:

Step 4. Beats in the wind (e). The direct sound now travels with the wind at $350\ \text{m/s}$, and the cyclist runs from it, so it passes the cyclist at $348\ \text{m/s}$: $f_d' = 510\ \text{Hz} \times \dfrac{348\ \text{m/s}}{350\ \text{m/s}} = 507.09\ \text{Hz}$. The beats are $513.09\ \text{Hz} - 507.09\ \text{Hz} = 6.01\ \text{Hz}$.