(b) When two similar bulbs of different powers are connected in series, then
$\frac{1}{P} = \frac{1}{P_{1}} + \frac{1}{P_{2}}$
Given, $P_{1} = 200\ W,\ P_{2} = 100\ W$
$\Rightarrow \frac{1}{P} = \frac{1 + 2}{200}$
$\Rightarrow P = \frac{200}{3} = 66.7\ W$