Answer
The 40-W bulb will glow brighter.
The brightness will be independent of the order since they are connected in series.
None of the bulbs are at full power.
Work Step by Step
R of 60-W bulb $=R_{60}=\frac{120^{2}}{60}=240ohm$
R of 40-W bulb $=R_{40}=\frac{120^{2}}{40}=360ohm$
So, total resistance when they are in series = $240+360=600ohm$
Thus, $I=\frac{120}{600}=0.2A$
Power consumed by 60-W bulb = $P_{60}=I^{2}R=0.2^{2}\times 240=9.6W$
Power consumed by 40-W bulb = $P_{40}=I^{2}R=0.2^{2}\times 360=14.4W$
Therefore the 40-W bulb will glow brighter. The brightness will be independent of the order since they are connected in series.
None of the bulbs are at full power.
