Two bulbs consume same energy when operated at 200 V and 300 V, respectively. When these bulbs are connected in series across a dc source of 500 V, then

$\begin{array}{l}\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\text{ or }\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}\text{ or }\mathrm{R}\propto {\mathrm{V}}^2,\\ \text{ i.e., }\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}={\left(\frac{200}{300}\right)}^2=\frac{4}{9}\end{array}$

When connected in series, potential drop is in the ratio of their resistances. So,

$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{4}{9}$

Now, $\mathrm{P}={\mathrm{I}}^2\mathrm{R}$

or  $\mathrm{P}\propto \mathrm{R}$ (in series I is the same)

or  $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{4}{9}$