How does the vapor pressure in a vessel affects the total pressure after a combustion reaction?
First, kudos for actually showing work on a problem!

Second, I'd solve this by inspection, rather than using the Ideal Gas Law, where time is critical on a test.

As you correctly calculated, the limiting reactant is $\ce{H2}$. Therefore the pressure of $\ce{H2}$ is nil; all the $\ce{H2}$ is in the water.

Half of the $\ce{O2}$ is used, reacting with the $\ce{H2}$ to make $\ce{H2O}$, so the remaining partial pressure of $\ce{O2}$ is ~157 torr.

The volume taken up by the liquid water is ~1/2,000 that of the original gases, so the volume left in the flask is still ~3 l, though slightly less. Since the answer choices are not close together, and the decrease in volume is negligible, I'd ignore it.

The vapor pressure of water at 25° C is ~24 torr, and at 30° C it's ~32 torr, so at 27° C, it would be about halfway, by linear interpolation, say ~28 torr.

The sum of those partial pressures is close to an answer choice.

Trivia: This question seems to be citing pi with the initial value, 3[.]14 torr. If the temperature had been -8° C (ice at that temperature has a v.p. ~2 torr), the answer would have been 159 torr, so the first six digits of pi could have been used; i.e. 314 ->159.