$P=\frac{V^2}{R}$

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