When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be :-

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.