Resistance of a bulb
$R=\frac{V^{2}}{P}$
$R_{1}=\frac{(220)^{2}}{25}=1936 \Omega$
$R_{2}=\frac{(220)^{2}}{100}=484 \Omega$
Since, $R_{1}$ and $R_{2}$ are in series
$$
\begin{aligned}
R_{n e t} &=R_{1}+R_{2} \\
&=1936+484=2420 \Omega \\
\text { rent } I &=\frac{V}{R}
\end{aligned}
$$
$$
\text { Hence, current } I=\frac{V}{R}
$$
$$
=\frac{440}{2420}=\frac{2}{11} \mathrm{~A}
$$
Potential difference across $25 \mathrm{~W}$ bulb
$$
\begin{aligned}
V_{1} &=I R_{1} \\
&=\frac{2}{11} \times 1936=352 \mathrm{~V}
\end{aligned}
$$
Potential difference across $100 \mathrm{~W}$ bulb
$$
V_{2}=I R_{2}=\frac{2}{11} \times 484=88 \mathrm{~V}
$$
Potential difference across $25 \mathrm{~W}$ bulb in this combination is $352 \mathrm{~V}$, but is can tolerate only $220 \mathrm{~V}$.
Hence, it will fuse.