a
In given Kettle $R=\rho \frac{ L }{\pi\left(\frac{ d }{2}\right)^2}=\frac{4 \rho L }{\pi d ^2}$
$P=\frac{V^2}{R}$
In second Kettle $R_1=\rho \frac{L}{\pi d^2}$
$R_2=\frac{\rho L}{\pi d^2}$
So $\quad R_1=R_2=\frac{R}{4}$
If wires are in parallel equivalent resistance
$R_P=\frac{R}{8}$
then power $P_P=8 P$
so it will take 0.5 minute
If wires are in series equivalent resistance
$R_s=\frac{R}{2}$
then power $P_s=2 P$ so it will take $2$ minutes