Answer
(C) $2.11\times 10^5 N$
Work Step by Step
We know that
$a=\frac{v^2-v_{\circ}^2}{2s}$
We plug in the known values to obtain:
$a=\frac{(0m/s)^2-(18m/s)^2}{2(0.05m)}$
$a=-3240m/s^2$
Now the force on the driver can be calculated as:
$F=(65Kg)(-3240m/s^2)$
$F=-2.11\times 10^5N$
Thus, the magnitude of force on the driver is $2.11\times 10^5N$ and hence the correct option is $C$.