not fuse $I_{1}=\frac{W_{1}}{V_{1}}=\frac{25}{220}\, \mathrm{Amp}$

Similarly, $\mathrm{I}_{2}=\frac{W_{2}}{V_{2}}=\frac{100}{220} \,\mathrm{Amp}$

The current flowing through the circuit

$I=\frac{440}{R_{e f f}}, \quad \mathrm{R}_{\mathrm{eff}}=R_{1}+R_{2}$

$R_{1}=\frac{V_{1}^{2}}{P_{1}}=\frac{(220)^{2}}{25} ; \quad R_{2}=\frac{V_{2}^{2}}{P}=\frac{(220)^{2}}{100}$

$I=\frac{440}{\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100}}=\frac{440}{(220)^{2}\left[\frac{1}{25}+\frac{1}{100}\right]}$

$I=\frac{40}{220}\, \mathrm{Amp}$

${I_1}\left( { = \frac{{25}}{{220}}A} \right) < I\left( { = \frac{{40}}{{220}}A} \right) < {I_2}\left( { = \frac{{100}}{{200}}A} \right)$

Thus the bulb marked $25\, \mathrm{W}-220$ will fuse