If three bulbs of 40 W, 60 W and 100 W are connected in series with a 200 V power supply, then

Power= $I^{2}R$

Since al three are connected in series the current flowing through all three bulbs is same.

Current flowing in the circuit is $\mathrm{i}=\frac{{\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_2}{\mathrm{Voltage} }=\frac{40+60+100}{200}=1\mathrm{A}$

Resistance of 40W , 60W and 100W are 40$\Omega$ , 60$\Omega$ and 100$\Omega$ respectively