1. Find the resistance of each bulb from its rating

For a bulb rated at voltage $V$ and power $P$, $R=\frac{V^2}{P}$

So,

• For the $100\,\text{W}$ bulb: $R_1=\frac{220^2}{100}=484\,\Omega$

• For the $60\,\text{W}$ bulb: $R_2=\frac{220^2}{60}=\frac{48400}{60}\approx 806.67\,\Omega$

2. Since the bulbs are in series, total resistance is $R_{\text{eq}}=R_1+R_2=484+806.67=1290.67\,\Omega$

3. Find the current through the series combination

Same current flows through both bulbs: $I=\frac{V}{R_{\text{eq}}}=\frac{220}{1290.67}\approx 0.1705\,\text{A}$

4. Power consumed by the 100 W bulb

Using $P=I^2R$, $P_1=I^2R_1=(0.1705)^2\times 484$ $P_1\approx 0.0291\times 484\approx 14.1\,\text{W}$

So the power consumed by the $100\,\text{W}$ bulb is about $\boxed{14\,\text{W}}$

5. Comparison with stored answer

Stored correct answer = $14$

Our derived answer = $14$

Hence, they agree.