1. Use the stopping distance relation

When the same braking force is applied, the retardation $a$ is constant. Using

$v^2=u^2+2as$

For stopping, final velocity $v=0$, so

$0=u^2+2as$ Rightarrow s=-\frac{u^2}{2a}$$

Thus, stopping distance is proportional to the square of initial speed:

$s\propto u^2$

2. Compare the two cases

Let the original speed be $u$, and stopping distance be $s_1=27\text{ m}$.

If the new speed is one-third of the original, then

$u_2=\frac{u}{3}$

So the new stopping distance is

$s_2=s_1\left(\frac{u_2}{u}\right)^2$

$s_2=27\left(\frac{1}{3}\right)^2=27\cdot\frac{1}{9}=3\text{ m}$

3. Final answer

The car will stop after travelling

$\boxed{3\text{ m}}$