Heating Effect of Current

In parallel, the potential drop V across both resistors is the same. $P = \frac{V^2}{R} \Rightarrow P \propto \frac{1}{R}$ $\frac{P_1}{P_2} = \frac{R_2}{R_1} = \frac{200}{100} = \frac{2}{1}$ So the ratio of thermal energies is 2 : 1.

Heat required to boil the water: $Q = mc\,\Delta T$
$Q = 1 \times 1 \times (100 - 20) = 80$ kcal
In joules, taking $c = 4200$ J kg$^{-1}$ K$^{-1}$: $Q = 1 \times 4200 \times 80 = 336000$ J
Heat given by the supply in time $t$: $H = V\,i\,t = 220 \times 4 \times t = 880\,t$ J
Setting $H = Q$:
$880\,t = 336000$
$t = \dfrac{336000}{880} = 381.8$ s
$t = \dfrac{381.8}{60} \approx 6.3$ min
So the kettle takes about 6.3 min to boil the water.

The same quantity of heat Q is needed to boil the water each time, and the supply voltage V is the same.
$Q = \dfrac{V^2}{R_1}t_1 = \dfrac{V^2}{R_2}t_2 = \dfrac{V^2}{R}t$, where R is the resistance of the parallel combination.

For parallel coils: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$
From the heat equation, $\dfrac{1}{R} = \dfrac{Q}{V^2 t}$, so $\dfrac{Q}{V^2t} = \dfrac{Q}{V^2t_1} + \dfrac{Q}{V^2t_2}$
$\Rightarrow \dfrac{1}{t} = \dfrac{1}{t_1} + \dfrac{1}{t_2} \Rightarrow t = \dfrac{t_1t_2}{t_1 + t_2}$
$t = \dfrac{10 \times 40}{10 + 40} = 8$ min