What is the torque of the force F=(2i−3j+4k)N acting at the point r=(3i+2j+3k)m about the origin?

For the cube to begin to topple, the torque due to the applied force $F$ about the edge must be equal to or greater than the torque due to the weight of the cube about the same edge. The pivot point for toppling will be the edge closest to the applied force $F$.

The weight of the cube, $mg$, acts downwards through its center of mass, which is at a height $\frac{a}{2}$ from the base and at a distance $\frac{a}{2}$ from the edge.
The torque due to weight about the pivot edge is $\tau_{mg} = mg \times \frac{a}{2}$.
The horizontal force $F$ is applied at a height $\frac{3a}{4}$ above the base.
The torque due to the force $F$ about the pivot edge is $\tau_F = F \times \frac{3a}{4}$.

At the point of toppling, the torque due to $F$ must overcome the torque due to $mg$.

The minimum value of $F$ for which the cube begins to topple is $\frac{2mg}{3}$.

Comparing this with the given options:

1. $\frac{3mg}{4}$
2. $\frac{2mg}{4}$
3. $\frac{2mg}{3}$
4. None of the above

The calculated value matches option 3.

The final answer is $\text{3. } \frac{2mg}{3}$.