An electrical bulb rated 220V, 100W , is connected in series with...

Electric power is given by $P = \dfrac{{V}^{2}}{R}$, here, $V$ is voltage and $R$ is resistance.

Now, for each bulb, ${R}_{1} = \dfrac{{V}^{2}}{P} = \dfrac{{220}^{2}}{100} = 484Ω$ and ${R}_{2} = \dfrac{{V}^{2}}{P} = \dfrac{{220}^{2}}{60} = 484\left(\dfrac{10}{6}\right)Ω$

Net current through the series combination is $I = \dfrac{220}{484 + 484 \times \dfrac{10}{6}}A = 0 . 171A$

Hence, the power consumed by the $100W$ bulb is ${P}_{1} = {I}^{2}{R}_{1} = 14 . 06W$