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 $N = 6$.