Since the anode & cathode both have the same voltages, the Eo of the cell = 0.
Now, using the Nernst equation, we can solve for the [Ag+].
Ecell = Eocell - (0.0592/n) log {[concentrated]/[dilute]} where n = 1 (Ag --> Ag+ + 1 e-) and [concentrated] = [1.00 M]. As stated above, Eocell = 0
0.031 = - 0.0592/1 log [1.00]/[Ag+]
Solving for [Ag+] gives us 0.524 V