Enthalpy of combustion is the change in enthalpy. The enthalpy change of a combustion reaction is called ΔH (combustion). The heat supplied to the calorimeter represents the heat supplied by virtue of the reaction taking place.
Normally, ΔH = ΔU + PΔv
The calorimeter is constant volume, so ΔV = 0. Hence, ΔH = ΔU in this case. So, as per the given conditions, for this particular reaction, the enthalpy of combustion and the internal energy change of combustion are the same.
So, usually when a problem like this is given, the heat absorbed by the calorimeter is ΔH, but here ΔH = ΔU given that no volume change occurs.
In your solution, you have made a pretty big mistake.
So, normally what we do is, we write ΔH = ΔU + PΔV, right? And PΔV = nRΔT. But this is for the gas! What you have done is you have taken the calorimeter to be an ideal gas! Hence ΔH = ΔU + nRΔT is applicable to the ΔT of the gas, not to the ΔT of the calorimeter. (On a side note, you are making an ideal gas assumption by assuming PΔV = nRΔT to be valid, which we cannot be sure of).
-> Heat absorbed by calorimeter represents the enthalpy change of the reaction
-> ΔH = ΔU + PΔV.
ΔV = 0, so ΔH = ΔU
-> PΔV = nRΔT is applicable to ideal gases only! Do NOT take the ΔT of the calorimeter as we don't have any data about temperature changes of the gases inside.
-> I think for most IIT-JEE problems, the ideal gas assumption is valid. But use PΔV first, don't jump to nRΔT without thinking.
-> Think about which ΔT you're using and whose temperature is changing.