Resistance of bulb is inversely proportional to its rated power.

$\Rightarrow R=\frac{V_{\text {rated }}^2}{P_{\text {rated }}} \Rightarrow R \propto \frac{1}{P_{\text {rated }}}$

As bulbs are in series, so same current flows through them at all instances. Power dissipation in a series combination is

$P=I^2 R \Rightarrow P \propto R$

$\text { or }P \propto \frac{1}{P_{\text {rated. }}}$

So, power dissipation is more in the $100 W$ bulb. This makes option (d) correct.