Answer
$F = 13.6~N$
Work Step by Step
Let the rotation axis be the point where the wheel touches the corner of the step.
The clockwise torque from the applied force must be enough to overcome the counter-clockwise torque from the weight of the wheel.
We can find the horizontal distance between the center of mass of the wheel and the rotation axis:
$x = \sqrt{r^2-(r-h)^2}$
$x = \sqrt{r^2-(r^2-2rh+h^2)}$
$x = \sqrt{2rh-h^2}$
$x = \sqrt{(2)(6.00~cm)(3.00~cm)-(3.00~cm)^2}$
$x = \sqrt{27}~cm$
We can find the minimum required force:
$\sum \tau_i = 0$
$(\sqrt{27}~cm)mg-(r-h)~F = 0$
$(r-h)~F = (\sqrt{27}~cm)mg$
$F = \frac{(\sqrt{27}~cm)mg}{r-h}$
$F = \frac{(\sqrt{27}~cm)(0.800~kg)(9.8~m/s^2)}{6.00~cm-3.00~cm}$
$F = 13.6~N$
