When the speed of a car is ‘u’, the minimum distance over which it can be stopped is ‘s’. If the speed becomes ‘nu’, what will be the minimum distance over which it can be stopped during the same time? 

As it is mentioned it has to be stopped in the same time the acceleration of the particle must also be changed 

In the first situation $a=\frac{u}{t}$

So in the second situation $a\text{'}=\frac{nu}{t}=na$

 ‘na’ assuming the acceleration in initial situation was ‘a’.

$$\begin{gathered} for first situation 0=u^2-2as \Rightarrow a=\frac{u^2}{2s} \\ so in the second situation, 0={\left(nu\right)}^2-2.\left(n\frac{u^2}{2s}\right).s\text{'} \\ s\text{'}=ns \end{gathered}$$