Electric Power and Combination of Bulbs

When the rated voltage of each bulb equals the supply voltage, the resistance of each bulb is $R = \dfrac{V^2}{P}$, and this resistance stays the same in any combination.
For bulbs connected in series across the same supply, the resistances add: $R_{eq} = R_1 + R_2 + R_3 + \dots$
Since $P = \dfrac{V^2}{R}$, i.e. $R \propto \dfrac{1}{P}$ for a fixed supply voltage, the effective power of a series combination is given by
$\dfrac{1}{P_{eq}} = \dfrac{1}{P_1} + \dfrac{1}{P_2} + \dfrac{1}{P_3} + \dots$
For this question: $\dfrac{1}{P_{eq}} = \dfrac{1}{60} + \dfrac{1}{60} + \dfrac{1}{60} = \dfrac{3}{60} = \dfrac{1}{20}$
$\Rightarrow P_{eq} = 20$ watt
Check: $R = \dfrac{200^2}{60} = 666.7\ \Omega$ for each bulb, $R_{eq} = 2000\ \Omega$, so $P = \dfrac{200^2}{2000} = 20$ W.
So the power drawn by the three bulbs is 20 watt.

The rated voltage equals the supply voltage, so we can use:
Series connection: $\dfrac{1}{P} = \dfrac{1}{P_1} + \dfrac{1}{P_2}$
Parallel connection: $P = P_1 + P_2$
Series: $\dfrac{1}{P} = \dfrac{1}{100} + \dfrac{1}{100} = \dfrac{1}{50} \Rightarrow P = 50$ W
Parallel: $P = 100 + 100 = 200$ W
Check: each bulb has $R = \dfrac{220^2}{100} = 484\ \Omega$; in series $\dfrac{220^2}{968} = 50$ W, and in parallel $\dfrac{220^2}{242} = 200$ W.
So the power drawn is 50 watt and 200 watt respectively.