A filament bulb ( 500W, 100V ) is to be used in a 230V main supply....

In the series, the current is the same in the bulb and the extra resistance, which is given by,

$I = \dfrac{P}{V} = \dfrac{500}{100} = 5A$.

But, ${V}_{bulb} + {V}_{R} = {V}_{net}$.

So, voltage across resistance $R$ will be,

${V}_{R} = {V}_{net} - {V}_{bulb}$
$⇒ {V}_{R} = 230 - 100 = 130V$.

From Ohm's law,

$R = \dfrac{{V}_{R}}{I} = \dfrac{130}{5} = 26Ω$.