The car will travel up to a distance of 146.28 m, hence the correct answer is 146 m.
Since the car is decelerating with the constant acceleration, so we can apply the third equation of motion.
v^2=u^2-2aS
here, v is the final speed of the car, which is 0 as the car stops,
u is the initial speed=32 m/s
a is the constant acceleration=3.5 m/s^2
and S is the distance cover by the car before it stops.
Now plugging the values in the third equation of motion
0=32^2-(2*3.5*S)
S=146.28 m
Therefore the car will cover a distance of 146 m before it stops.