Here is my response to you earlier question. If you still have questions you may want to follow up here.

Question

You didn't substitute correctly for Keq. You should calculate
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
You made log Keq = 0.27. You should follow the instructions I gave.
(Pb) = 1
(Ni^2+) = 0.27
(Ni) = 1
(Pb^2+) = x so the equation is
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)] and solve for x in moles/L.

I don't want to leave without acknowledging an error in what I wrote. It make no difference in the final answer becasue I corrected it later. Here is what I wrote for ONE of the steps.
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) but I omitted Faraday's constant. It should be
Ecell = Eo cell - (2.303*RT/nF) log Q and Q = Keq=(Pb)(Ni^2+)/(Ni)(Pb^2+)
Then if you substitute for R of 8.314, T = 298, F = 96,485 and 2 for n, then
(2.303*8.314*298/2*96,485) = 0.0296. Your post is now down below the line. You may want to repost at the top if you have additional quesitons.