Given that, initial speed of car, $v_i=v$
Final speed of car when it stops, $v_f=0$ Distance travelled $=s$
Retarding force $=F$
By Newton's law of motion,
$$
\begin{aligned}
F=m a & =m\left(\frac{v_f^2-v_i^2}{2 s}\right) \quad \ldots \text { (i) }\left[\begin{array}{l}
\text { using relation } \\
a=\frac{v_f^2-v_i^2}{2 s}
\end{array}\right] \\
& =\frac{m\left(-v^2\right)}{2 s} \\
F & =\frac{-m v^2}{2 s}
\end{aligned}
$$
When, retarding force, $F^{\prime}=3 F$
Let $s^{\prime}=$ New distance travelled.
Similarly, from Eq. (i) we can write
$$
3 F=\frac{-m v^2}{2 s^{\prime}}...(ii)
$$
Substituting value of $F$ from Eq. (i) into Eq. (ii), we get
$$
3\left(\frac{-m v^2}{2 s}\right)=\frac{-m v^2}{2 s^{\prime}}
$$
$$
\Rightarrow
$$
$$
s^{\prime}=s / 3
$$
Hence, the body will stop by travelling distance $\frac{s}{3}$, when retarding force is $3 F$.