The moment of a force about a point (also called torque) is calculated as the product of the force and the perpendicular distance from the point to the line of action of the force.

The force is $F_1 = 50 \text{ N}$.
The perpendicular distance from point O to the $50 \text{ N}$ force is $d_1 = 3 \text{ m}$.

$$\text{Moment}_{50 \text{ N}} = F_1 \times d_1 = 50 \text{ N} \times 3 \text{ m}$$ $$\text{Moment}_{50 \text{ N}} = 150 \text{ Nm}$$

The force is $F_2 = 80 \text{ N}$.
The perpendicular distance from point O to the $80 \text{ N}$ force is $d_2 = 4 \text{ m}$.

$$\text{Moment}_{80 \text{ N}} = F_2 \times d_2 = 80 \text{ N} \times 4 \text{ m}$$ $$\text{Moment}_{80 \text{ N}} = 320 \text{ Nm}$$

• Moment of $50 \text{ N}$ is $150 \text{ Nm}$.
• Moment of $80 \text{ N}$ is $320 \text{ Nm}$.

This matches option (C).

The final answer is $\boxed{\text{C) moment of } 50 \text{ N is } 150 \text{ Nm and } 80 \text{ N is } 320 \text{ Nm}}$.