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

We know that $\mathrm{R}={\mathrm{V}}^2/\mathrm{W}$
$\therefore$ Resistance of 40 W bulb
$=\frac{200\times 200}{40}=1000\mathrm{\Omega }$
Resistance of 60 W bulb
$=\frac{200\times 200}{60}=666\cdot 7\mathrm{\Omega }$
Resistance of 100 w bulb
$=\frac{200\times 200}{100}=400\mathrm{\Omega }$
Total resistance of the circuit
$=1000+666\cdot 7+400=2066\cdot 7\mathrm{\Omega }$
Current in the circuit
$=\frac{200}{2066\cdot 7}=0\cdot 097\mathrm{amp}$
As the bulbs are connected in series and hence the same current (0.097 amp.) flows through each bulb. The potential difference is expressed by V = iR. So, the potential difference across 40 W bulb would be maximum because its resistance is highest and current through all bulbs is the same.