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

Now, for each bulb, $R_1=\frac{V^2}{P}=\frac{{220}^2}{100}=484 \mathrm{\Omega }$ and  $R_2=\frac{V^2}{P}=\frac{{220}^2}{60}=484\left(\frac{10}{6}\right) \mathrm{\Omega }$

Net current through the series combination is $I=\frac{220}{484+484\times \frac{10}{6}} \mathrm{A}=0.171 \mathrm{A}$

Hence, the power consumed by the $100 \mathrm{W}$ bulb is $P_1=I^2{R}_1=14.06 \mathrm{W}$