$\frac{(3 E)^{2}}{R} t=(m) s \Delta T$$.......................(i)$
When the length of the wire is doubled, resistance and mass both are doubled. Therefore, in the second case
$\frac{(N E)^{2}}{2 R} t=(2 m) s \Delta T$
Dividing Eq. $(ii)$ by Eq. $( i ),$ we get $N=6$




