Answering the question from the title directly:

The bulb with lower power rating has a higher resistance (at operating temperature, but we have to assume it to be approximately constant), because at nominal voltage, less current must flow through it compared to the bulb with higher rating.

In a series circuit, the current is equal at all points.

The voltage drop at each bulb can be calculated using $V = RI$ (assuming constant $R$ despite changing temperature of the filament). Higher resistance leads to higer voltage drop at the bulb with lower wattage.

Power is calculated using $P=VI$, and since $I$ is constant for both bulbs, higher resistance leads to higer power, more brightness.

Answering the question from the title directly:

The bulb with lower power rating has a higher resistance (at operating temperature, but we have to assume it to be approximately constant), because at nominal voltage, less current must flow through it compared to the bulb with higher rating.

In a series circuit, the current is equal at all points.

The following assumes constant $R$ for each bulb.

The voltage drop at each bulb can be calculated using $V = RI$ (assuming constant $R$ despite changing temperature of the filament). Higher resistance leads to higer voltage drop at the bulb with lower wattage.

For 60W @ 120V we need 0.5A of current and with $R=V/I$ we get $R_{60W} = 240\Omega$. The other bulb has $R_{100W}= 144\Omega$.

Power is calculated using $P=VI$, and since $I$ is constant for both bulbs, higher resistance leads to higer power, more brightness.

The current is $I=V / (R_{60W} + R_{100W}) = 120V / 384 \Omega$