1. Frequency reflected from a moving car

When a stationary source emits sound of frequency $f_0$ toward a car approaching it with speed $u$, the car first acts as an observer and then as a moving source after reflection.

So the reflected frequency is

$f_r = f_0\left(\frac{v+u}{v-u}\right)$

where $v=330\,\text{m s}^{-1}$ is the speed of sound.

2. For two cars with speeds $u_1$ and $u_2$

The reflected frequencies are

$f_1=f_0\left(\frac{v+u_1}{v-u_1}\right), \qquad f_2=f_0\left(\frac{v+u_2}{v-u_2}\right)$

Given that the difference between these reflected frequencies is $1.2\%$ of $f_0$:

$|f_1-f_2|=0.012f_0$

So,

$\left|\frac{v+u_1}{v-u_1}-\frac{v+u_2}{v-u_2}\right|=0.012$

3. Use the approximation $u\ll v$

Since the car speeds are much smaller than the speed of sound,

$\frac{v+u}{v-u} \approx 1+\frac{2u}{v}$

Thus,

$f_r \approx f_0\left(1+\frac{2u}{v}\right)$

Hence,

$|f_1-f_2| \approx f_0\left|\frac{2u_1}{v}-\frac{2u_2}{v}\right| = f_0\frac{2|u_1-u_2|}{v}$

Given this equals $0.012f_0$,

$\frac{2|u_1-u_2|}{v}=0.012$

$|u_1-u_2|=\frac{0.012\times v}{2}$

Substitute $v=330\,\text{m s}^{-1}$:

$|u_1-u_2|=\frac{0.012\times 330}{2}=1.98\,\text{m s}^{-1}$

4. Convert to km/h

$1.98\times 3.6 = 7.128\,\text{km h}^{-1}$

To the nearest integer,

$\boxed{7}$

5. Comparison with stored answer

Derived answer = $7$

Stored correct answer = $7$

They agree.