The copper nucleus and the proton we're bringing in will repel each other. The electric force will point in the opposite direction to us moving the proton toward the nucleus, so a net positive amount of external work will have to be done against the electric field. Thus the electric field does a net negative amount of work.
The work we have to do against the field is given by:
W = ΔVq
W = work, ΔV = potential difference, q = proton charge
If the 0 potential is defined to be infinitely far away, then the potential difference ΔV is given by:
ΔV = kQ/r
Q = nucleus charge, r = radius of atom
Make a substitution:
W = kQq/r
Given values:
k = Coulomb constant, Q = 29e (elementary charges), q = 1e, r = 4.8×10⁻¹⁵m
Plug in and solve for W:
W = (9×10⁹)(29)(1)/(4.8×10⁻¹⁵)
W = 8.7MeV
Put a negative sign in front of W, and now we have the work done by the electric field, -8.7MeV