To solve the problem it is necessary to apply the concepts related to torque, as well as the concepts where the Force is defined as a function of mass and acceleration, which in this case is gravity.
Considering the system in equilibrium, we perform sum of moments in the rear wheel (R2)
[tex]\sum M = 0[/tex]
[tex]F_g*1.68-R_1*d = 0[/tex]
[tex]mg*1.68-R_1*d = 0[/tex]
Another of the parameters given in the problem is that the front wheel supports 43% of the weight, that is
[tex]R1=0.43F_g[/tex]
[tex]R1=0.43mg[/tex]
Replacing, we have to
[tex]mg*1.68 -R_1*d = 0[/tex]
[tex]mg*1.68 -0.43mg*d = 0[/tex]
[tex]mg*1.68 =0.43mg*d[/tex]
[tex]1.68 = 0.43*d[/tex]
[tex]d =3.9m[/tex]
Therefore the wheelbase of the truck is 3.9m between the front and the rear.