Let car B be the observer (moving towards S).

The frequency observed is

$${f_1} = {f_0}\left( {{{c + v} \over c}} \right)$$

When sound gets reflected, the frequency observed by source S is

$${f_2} = {f_1}\left( {{c \over {c - v}}} \right)$$

where v is the speed of car and c is the speed of sound. Therefore,

$${f_2} = {f_0}\left( {{{c + v} \over {c - v}}} \right)$$

Now, $$d{f_x} = {f_0}\left[ {{{(c - v)dv - (c + v)( - dv)} \over {{{(c - v)}^2}}}} \right]$$

$$ = {{2{f_0}c\,dv} \over {{{(c - v)}^2}}}$$

That is,

$${{2{f_0}c\,dv} \over {{{(c - v)}^2}}} = \left( {{{1.2} \over {100}}} \right){f_0}$$

$$ \Rightarrow dv = {{1.2} \over {100}} \times {{{{(c - v)}^2}} \over {2c}}$$

Since, v << c, we get c $$-$$ v $$ \simeq $$ c.

Therefore,

$$dv = {{1.2} \over {100}} \times {c \over 2} = 1.98$$ m/s

$$ = 1.98 \times {{18} \over 5}$$ km/h = 7 km/h.