Here we will use the formula of power P=RV2 . So resistance R=PV2 . Resistance of 25W bulb R1=25(220)2=1936ohm and resistance of 100W bulb R2=100(220)2=484ohm . When they are connected in series with a supply of 440V , the resultant current will be I=(1936+484)440=0.1819A
Now the voltage drop across 25W bulb will be V1=IR=0.1819×1936=352V and the voltage drop across 100W bulb will be V2=0.1819×484=88V . Here because of the excessive voltage (much higher than rated value) across 25W bulb, it will fuse.