Explanation

(b) The power of 1$^{st}$ bulb
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F_{1} = \frac{V^{2}}{R_{1}} = \frac{(220)^{2}}{R_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ 100 = \frac{(220)^{2}}{R_{1}}$
or $R_{1} = \frac{200\ \times 200}{100} = 400\ \Omega$
The power of IInd bulb
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{2} = \frac{(200)^{2}}{R_{2}}$
or $\ \ \ \ \ \ \ 200 = \frac{(200)^{2}}{R_{2}}$
or $\ \ \ \ \ \ \ \ \ \ R_{2} = \frac{200\ \times 200}{200} = 200\ \Omega$
The bulbs are joined in series.
So, $R = R_{1} + R_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 400 + 200 = 600\ \Omega$
The total power
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P = \frac{V^{2}}{R} = \frac{(200)^{2}}{600}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P = 66.7\ W$