when the bulbs are connected in series

${{\rm{R}}_{\rm{s}}} = {{\rm{R}}_{{{\rm{B}}_1}}} + {{\rm{R}}_{{{\rm{B}}_2}}} = \frac{{{{(220)}^2}}}{{40}} + \frac{{(220)}}{{60}}$

${=(220)^{2}\left[\frac{1}{40}+\frac{1}{60}\right]=(220)^{2}\left(\frac{100}{2400}\right)=\frac{(220)^{2}}{24}}$

$\therefore \quad \mathrm{P}_{1}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{2}}=(220)^{2} \times \frac{24}{(220)^{2}}=24 \mathrm{\,W}$

$P_{2}=60+40=100 \mathrm{\,W}$