(A) Mass of the automobile,$m = 3000 \; kg$.
Displacement in the suspension system,$x = 15 \; cm = 0.15 \; m$.
There are $4$ springs in parallel to support the mass of the automobile.
The equation for the restoring force is $F = 4kx = mg$.
Thus,the spring constant $k$ for one wheel is $k = \frac{mg}{4x} = \frac{3000 \times 9.8}{4 \times 0.15} = 49000 \; N/m = 4.9 \times 10^4 \; N/m$.
For one wheel,mass $M = 750 \; kg$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{750}{49000}} \approx 0.778 \; s$.
The amplitude of oscillation decreases by $50\%$,so $A = A_0 e^{-bt/2M} = 0.5 A_0$.
Taking natural log,$\ln(2) = \frac{bt}{2M}$.
$b = \frac{2M \ln(2)}{T} = \frac{2 \times 750 \times 0.693}{0.778} \approx 1336.5 \; kg/s$.