Calculating change in enthalpy of combustion of 1.12g of hexane I will calculate the change in enthalpy of combustion of 1.12g of hexane.

1) Calculate the energy transferred to 200g of water using the equation: J = mass of water in grams (200g) x specific heat capacity of water (4.18 J/gK) x temperature increase in Kelvin (24C = 297.15K). This results in the amount of energy transferred to the water of a total of 248,417.4J or 248.4kJ.

2) Calculate number of moles of hexane burnt using n=m/M. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore 1.15g of hexane was burnt during the experiment. 1.15g of hexane = 0.013mol of hexane.

3) Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) -4194 kJ/mol.  Therefore, 0.013mol will produce (0.013/-4194) -3.1x10-6 kJ/mol.

So this is the procedure that I used to find out the answer! Was it correct?
Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?
I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:

1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ 

2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol 

3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol

So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane.