Answer
$k = 2600~N/m$
Work Step by Step
$v = (95~km/h)(\frac{1000 ~m}{1~km})(\frac{1~h}{3600~s}) = 26.4~m/s$
The force exerted by a spring increases as the spring is compressed. Since the acceleration is proportional to the force, the car will experience the greatest acceleration at the point of maximum compression $x$.
We can find an expression for the maximum compression $x$. Let $a = 4.0~g$. Then,
$\sum F = ma$
$kx = 4.0~mg$
$x = \frac{4.0~mg}{k}$
We can use this expression for $x$ to find the spring constant $k$. At the point of maximum compression $x$, the elastic potential energy stored in the spring must equal the original kinetic energy of the car:
$KE = EPE$
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$
$mv^2 = (k)(\frac{4.0~mg}{k})^2$
$k = \frac{16.0~mg^2}{v^2} = \frac{(16.0)(1200~kg)(9.80~m/s^2)^2}{(26.4~m/s)^2}$
$k = 2600~N/m$