A boy riding on his bike is travelling towards the east at a speed...

The velocity of the source is $v_s = 4\sqrt{2} \text{ m s}^{-1}$ towards the east.

The observer is stationary, so $v_o = 0$. The position of the observer relative to the source is $45^\circ$ south of east.

The component of the source's velocity along the line joining the source and the observer is:

$v_{s,\text{eff}} = v_s \cos(45^\circ) = 4\sqrt{2} \times \dfrac{1}{\sqrt{2}} = 4 \text{ m s}^{-1}$

Using the Doppler effect formula for a stationary observer and a moving source, the apparent frequency $f'$ is given by:

f' = f_0 \left(\dfrac{v}{v - v_{s,\text{eff}}} \right)

Substituting the given values $f_0 = 1650 \text{ Hz}$ and $v = 334 \text{ m s}^{-1}$:

$f' = 1650 \left( \dfrac{334}{334 - 4} \right)$

f' = 1650 \left(\dfrac{334}{330} \right)

$f' = 5 \times 334 = 1670 \text{ Hz}$