Answer : The correct option is, (B) -5448 kJ/mol
Explanation :
First we have to calculate the heat required by water.

where,
q = heat required by water = ?
m = mass of water = 250 g
c = specific heat capacity of water = 
= initial temperature of water = 293.0 K
= final temperature of water = 371.2 K
Now put all the given values in the above formula, we get:


Now we have to calculate the enthalpy of combustion of octane.

where,
= enthalpy of combustion of octane = ?
q = heat released = -81719 J
n = moles of octane = 0.015 moles
Now put all the given values in the above formula, we get:


Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.