additional resistance required to produce a potential drop
I'll assume that the requirement is really 1 mV/cm of voltage drop along the wire. Here's one way to get there:

1 mV/cm &times; 100 cm = 100 mV total drop required.

100 mV / 3 &Omega; = 33.3 mA of current needs to flow.

2 V cell / 33.3 mA = 60 &Omega; total resistance is required.

60 &Omega; total - 3 &Omega; wire = 57 &Omega; additional required.