Since power P is given by P=V2/R, so R=V2/P
For the first bulb, R1=(P1V2)=[25(220)2]=1936Ω
For the second bulb, R2=(P2V2)=[100(220)2]=484Ω
Current in series combination is the same in the two bulbs and current i is given by i=R1+R2V=1936+484220 =2420220=111A
If the actual powers in the two bulbs be P1 and P2 then P1′=i2R1=(111)2×1936=16W
and P2′=i2R2=(111)2×484=4W
Since P1′>P2′,25W bulb will glow more brightly.