b
Let \(R\) and \(m\) be the resistance and mass of the first wire, then the second wire has resistance \(2R\) and mass \(2\,m\). 

Let \(E = emf\) of each cell, \(S =\) specific heat capacity of the material of the wire. For the first wire, current 

\({i_1} = \frac{{3E}}{R}\) and \(i_1^2Rt = mS\Delta T\)

For the second wire, \({i_2} = \frac{{NE}}{{2R}}\) and \(i_2^2(2R)t = 2\,mS\Delta T\). Thus, \({i_1} = {i_2}\) or \(N = 6\).