Answer
(a) If the travelling speed is doubled, the stopping distance will be $4x$.
(b) $x=6.8\,\mathrm{m}$
Work Step by Step
(a) It is given that $\Delta x=x$ and $v_i=v$.
Since the car comes to a stop, $v_f=0$ and $a$ is negative.
Using the kinematic equation
\begin{align*}
v_f^2&=v_i^2+2a\Delta x\\
0^2&=v^2-2ax\\
x&={v^2\over 2a}
\end{align*}
If the initial speed $v'$ is doubled. i.e., $v'=2v$, then the new stopping distance
$$x'={v'^2\over 2a}={4v^2\over 2a}=4x$$
The stopping distance quadruples if the initial speed is doubled.
(b) If $x=3.00\,\mathrm{m}$ and $v=40.0\,\mathrm{km/h}=11.1\,\mathrm{m/s}$, the acceleration is
$$a={v^2\over 2x}={11.1^2\over 2(3.00)}={123\over 6.00}=20.5\,\mathrm{m/s^2}$$
Now, if $v=60.0\,\mathrm{km/h}=16.7\,\mathrm{m/s}$, the modified stopping distance will be
$$x={v^2\over 2a}={16.7^2\over 2(20.5)}={279\over 41}=6.8\,\mathrm{m}$$
