Firstly, we need to understand some basic properties and their values. The charge of one electron is given by -1.6 x 10^(-19) coulombs. In our question, we are looking at a particle with 2 excess electrons, so the total charge for this particle, denoted as 'q', is 2 times the charge of one electron, which is -3.2 x 10^(-19) coulombs.
The magnitude of the electric field, denoted as 'E', is given as 15.1kN/C. As we know that 1 kN/C equals to 10^3 N/C, we need to convert it to N/C by multiplying it by 10^3. So, E = 15.1 x 10^3 N/C.
The formula to calculate the electric force 'F' acting on a charge in the presence of an electric field is F = qE, where q is the charge and E is the magnitude of the electric field. Substituting the values of q and E into this formula, we will get the magnitude of the electric force:
F = (-3.2 x 10^(-19) C) * (15.1 x 10^3 N/C)
This calculation will give us the electric force as: F = -4.832 x 10^(-15) N
Therefore, the magnitude of the electric force acting on the particle with 2 excess electrons in the presence of a 15.1 kN/C field is 4.832 x 10^(-15) N. Here, the negative sign indicates the direction of the force, which is towards the source of the field due to the negative charge of the electrons.