At the point of fission, a nucleus with 96 protons is divided into two smaller spheres, each with 48 protons?

Question

At the point of fission, a nucleus with 96 protons is divided into two smaller spheres, each with 48 protons and a radius of 6.4*10^-15 m. What is the repulsive force pushing these two spheres apart?

Answer 1

F=constant q q/r^2 where q q are the charges and r=6.4*10^-15 m
You need to convert everything into the same Units.