Three rigid rods are joined to form an equilateral triangle ABC of side 1m.Three particles carrying charge 1C are placed at the vertices.The whole system is in equilibrium in an inertial frame.What is the magnitude of the net force on each of the charges?

This may not be a simple question as it looks at first.In fact,it is an MCQ type question with four answers given from where we are to select the correct one.

The answer is zero.
Since the system is in equilibrium,the charges should also be in equilibrium too.
Hence,net force on each charge is to be zero.And that is the answer.

But what about the Coulomb's force and all that?They are there.And here we must assume external forces of equal magnitude acting over the charges in appropriate direction so that no net force acts on the system.

It should not be surprising to assume external forces if we recall the Earnshaw's theorem.

I think my logic is OK.

Please respond and let me know if I am correct.

Your logic seems sound. I'm thinking the charges are held in place by the rigid rods.

Each charge will have a large coulomb force (I think 15.5 Giganewtons!) away from the centre of the triangle. But as you say, the system is specified as being in equilibrium, therefore there must be some counterforce to stop them accelerating away, and the rigid rods would appear to fit that purpose.

So, I think that each charge experiences a coulomb force of 15.5GN outward and a mechanical restraining force of 15.5GN inward, for a net force of zero.

I did not calculated the force,but you are correct.The charges should have been attached to the inner vertices of the triangle.