Two 1.0 kg masses are 4.0 m apart on a frictionless table. each has 1.0 μc of charge. what is the magnitude of the electric force on one of the masses?

The electrostatic forces between two charges are

   F  =  (9 x 10⁹) x (charge-1) x (charge-2) / (distance between them)² .

The charges in this question are each (1 x 10⁻⁶ Coulomb).

   Force = (9 x 10⁹) (1 x 10⁻⁶) (1 x 10⁻⁶)  /  (4.0)

             =  (9 x 10⁹)         (1 x 10⁻¹²)          /  (4.0)

             =               0.00225 Newton .

-- The force on each mass is the same.
Each mass is acted on by a force of 0.00225 newton,
directed either toward the other one or away from it.

-- If the charges both have the same sign, then the forces
are repulsive ... driving the masses apart.

-- If the charges have different signs, then the forces
are attractive ... drawing the masses together.

-- For as far as you've gone in copying this question ... asking
only for the forces between the charged objects ... you don't
need to know their masses, or about friction with the table, or
any of that stuff.  None of that has any effect on the forces
between the charged masses.

You'll need that information when you get down to the next part
of the question, where it's probably going to ask you for the
acceleration of each mass due to the electrostatic force on it.

You know the force on each mass now.  All you have to remember
is  F = (mass) x (acceleration) , and grind it out.

Don't you dare post the next part on !  You have everything
you need to do it yourself now, and you'll learn a lot more that way.