A car moving with a speed of 50km h ^ - 1 can be stopped by brakes...

The stopping distance of a car is

$s = \dfrac{{u}^{2}}{2a}$

Therefore, for the two situations given in the question, we can write

$\dfrac{{s}_{1}}{{s}_{2}} = {\left(\dfrac{{u}_{1}}{{u}_{2}}\right)}^{2} = \dfrac{1}{4}$

$\dfrac{6}{{s}_{2}} = \dfrac{1}{4} ⇒ {s}_{2} = 24m$