To solve this problem, we need to calculate the moment (torque) of the $80 \sqrt{3} \, \text{N}$ force about point $A$.

Step-by-Step Solution:

1. Identify the Force and Distance:

   • The force is $80 \sqrt{3} \, \text{N}$ acting downward at point $C$.
   • Point $A$ is the pivot point.
   • The distance from $A$ to $C$ can be found using points $A$, $B$, and $C$.

2. Calculate the Perpendicular Distance (Moment Arm):

   • From the diagram:
     • $AB = 8 \, \text{m}$ (horizontal distance from $A$ to $B$)
     • $BC = 4 \, \text{m}$ at an angle of $120^\circ$ with the horizontal.
   • The perpendicular distance from point $A$ to the line of action of the force is the projection of $AC$ in a direction perpendicular to the force.

3. Using Trigonometry:

   • The angle between $AB$ and $BC$ is $120^\circ$.
   • The perpendicular distance from $A$ to the line of force (let’s call it $d$) is given by: $d = AB + BC \cos(120^\circ)$
   • Since $\cos(120^\circ) = -\frac{1}{2}$: $d = 8 + 4 \left(-\frac{1}{2}\right) = 8 - 2 = 6 \, \text{m}$

4. Calculate the Moment:

   • Moment about $A$ is calculated by: $M = F \cdot d$
   • Substituting $F = 80 \sqrt{3} \, \text{N}$ and $d = 6 \, \text{m}$: $M = 80 \sqrt{3} \times 6 = 480 \sqrt{3} \, \text{N} \cdot \text{m}$
   • Since the force would cause a clockwise rotation about $A$, the moment is negative: $M = -480 \sqrt{3} \, \text{N} \cdot \text{m}$

Answer:

The correct option is: (b) $-800\sqrt{3}$.