Since power $P$ is given by $P=V^{2} / R$, so
$R=V^{2} / P$
For the first bulb,
$R_{1}=\left(\frac{V^{2}}{P_{1}}\right)=\left[\frac{(220)^{2}}{25}\right]=1936\, \Omega$
For the second bulb,
$R_{2}=\left(\frac{V^{2}}{P_{2}}\right)=\left[\frac{(220)^{2}}{100}\right]=484\, \Omega$
Current in series combination is the same in the two bulbs and current $i$ is given by
$i =\frac{V}{R_{1}+R_{2}}=\frac{220}{1936+484}$
$=\frac{220}{2420}=\frac{1}{11} A$
If the actual powers in the two bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}'=i^{2} R_{1}=\left(\frac{1}{11}\right)^{2} \times 1936=16\, W$
and $P_{2}'=i^{2} R_{2}=\left(\frac{1}{11}\right)^{2} \times 484=4\, W$
Since $P_{1}' >P_{2}', 25\, W$ bulb will glow more brightly.