You need to use the following equation that relates the distance, speed, acceleration and time such that:

`d = v_0*t + (at^2)/2`

Since the car starts from rest, the initial speed is `v_0 = 0`  and the time of 3 s is provided by the problem, hence, substituting these values in equation above yields:

`d = 9a/2`

You need to find the value of acceleration, hence, using the Newton's second law of motion yields:

`m*a = F_f`

`F_f ` represents the friction force

`F_f = mu*N = mu*m*g => m*a = mu*m*g => a = mu*g`

Since the problem provides the coefficient of static friction mu = 0.24 and considering `g = 9.8 m/s^2`  yields:

`a = 0.24*9.8 = 2.352 m/s^2`

Substituting `2.352 ` for a in equation`d = 9a/2`  yields:

`d = 9a/2 => d = 9*2.352/2 = 10.584 m` ~~ 11m

Hence, evaluating the maximum distance the truck can travel, without having the box slide, yields `d~~11 m` , thus, you may select the answer b.