Calculate the force $F$ that is applied horizontal at the axle of the wheel which is necessary to raise the wheel over the obstacle of height $0.4 \, \text{m}$. Radius of wheel is $1 \, \text{m}$ and mass = $10 \, \text{kg}$. $F$ is 

At the instant the wheel just starts to rise, it rotates about the corner of the obstacle. So taking moments about that corner, the reaction there gives no moment. The perpendicular distance of the horizontal force from the corner is the vertical separation: $$1 - 0.4 = 0.6$$ The perpendicular distance of the weight from the corner is the horizontal separation. From geometry of the radius to the corner, $$x = \sqrt{1^2 - 0.6^2} = 0.8$$ Balancing moments about the corner, $$F \times 0.6 = mg \times 0.8$$ With $m = 10$ and $g \approx 10$, $$F \times 0.6 = 100 \times 0.8 = 80$$ $$F = \frac{80}{0.6} = 133.3\mathrm{N}$$ This matches option $\text{D}$.