What happens to its kinetic energy, when the speed of car gets doubled?

The correct option is E It is quadrupled
Kinetic energy of car $K=\frac{1}{2}m{v}^2$

Now the speed of car is doubled i.e $v^{\prime }=2v$

$\therefore$ New kinetic energy $K^{\prime }=\frac{1}{2}m(v^{\prime }{)}^2=\frac{1}{2}m(2v{)}^2=4K$