I don't know the context of the problem but your 979 kJ/mol appears to be right if you have the number of significant figures correct. Does the problem say anything about "about 10.0 millimoles" methane? You have P, V, R, and T and you can calculate n. If I can calculate n I get something like 0.0112 and if the q is divided by 0.0112 instead of 0.01 to obtain per mol the number you get is much closer to 890 kJ/mol than if you use 0.01. Perhaps that 10.0 mmols is "close" and you are to use PV = nRT to solve for the real number of mmols. Here is the site I talked about a couple days ago where this experiment is delineated. There is nothing in the material about using PV = nRT BUT the instruction say to add a known amount of methane and I don't know how you would do that without using PV = nRT. I suspect you are to use PV = nRT to solve for the real number of mols CH4.
http://www.chm.davidson.edu/vce/calorimetry/heatofcombustionofmethane.html