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

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

But, $V_{\mathrm{bulb}}+V_{\mathrm{R}}=V_{\mathrm{net}}$.

So, voltage across resistance $R$ will be,

$V_{\mathrm{R}}=V_{\mathrm{net}}-V_{\mathrm{bulb}}$
$\Rightarrow V_{\mathrm{R}}=230-100=130\mathrm{ }\mathrm{V}$.

From Ohm's law,

$R=\frac{V_{\mathrm{R}}}{I}=\frac{130}{5}=26 \mathrm{\Omega }$.