Greetings! Remember that torque is [tex]\tau = \overline{r} \times \overline{F}[/tex], which is different than [tex]\tau = rF[/tex]. In other words, torque is the radius at which a force acts perpendicularly to the lever arm, times the magnitude of the force. Thus, for part A, note that the corner of the step will be exerting a force [tex]F_{min}[/tex] directly to the left in order to counteract the rightward force on the wheel's axle. If we extend this leftward [tex]F_{min}[/tex], we see that it acts perpendicularly to the wheel at a radius of [tex]R - h[/tex] (the moment arm is conveniently indicated by the vertical dotted line).
Therefore, for part A, [tex]\tau = \overline{r} \times \overline{F} = (R-h)F_{min}[/tex]. Try applying this same strategy to part B.