A traffic policeman stationed on a road blows a whistle that emits...

The frequency of the source is $f = 2.00 \text{ kHz}$.

The speed of the observer is $v_o = 36.0 \text{ km/h} = 36.0 \times \dfrac{5}{18} \text{ m/s} = 10.0 \text{ m/s}$.

The speed of sound is $v = 340 \text{ m/s}$ and the speed of the source is $v_s = 0$.

Using the Doppler effect formula for an observer approaching a stationary source:

f' = f \left(\dfrac{v + v_o}{v} \right)

Substituting the given values:

$f' = 2.00 \times \left( \dfrac{340 + 10}{340} \right)$

f' = 2.00 \times \left(\dfrac{350}{340} \right)

$f' = 2.00 \times$\dfrac{35}{34} \approx$2.058 \text{ kHz}$

Rounding to two decimal places, $f' \approx 2.06 \text{ kHz}.$