If I understood the question correctly, the volume will be #V = 69.7 L#
Basically, you are dealing with the combustion of benzene, #C_6H_6#. Starting from the balanced reaction,
#2C_6H_6 + 15O_2 -> 12CO_2 + 6H_2O#
you can see that the benzene to oxygen ratio is #2:15#; that is, for every 2 moles of benzene you need 15 moles of oxygen.
Your initial values are #0.5# moles for #C_6H_6# and #3.75# moles for #O_2#, which correspond to the balanced equation's ratio. This means that neither the benzene, nor the oxygen are limiting reactants.
So, for every #2# moles of #C_6H_6# you will produce #12# moles of #CO_2# and #6# moles of #H_2O#.
GIven that you have #0.5# moles to start with, you will end up with #3# moles of #CO_2# ( 1:6 ratio ) and #1.5# moles of #H_2O# ( 1:3 ratio).
Now, using the ideal gas law #PV = nRT# and knowing that you have a total of #3 + 1.5 = 4.5# moles of gas produced, you get
#V = (nRT)/P = (4.5 * 0.082 * (105+273))/2.0 = 69.7 L#
