Start by taking a look at the balanced chemical equation
#color(red)(2)C_6H_(6(g)) + 15O_(2(g)) -> color(green)(12)CO_(2(g)) + color(blue)(6)H_2O_((g))#
Notice the mole ratios that exist between benzene and your two products, carbon dioxide and water.
The #color(red)(2):color(green)(12)#, or #color(red)(1):color(green)(6)#, mole ratio that exists between benzene and carbon dioxide tells you that, regardless of how many moles of benzene react, the reaction will produce 6 time more moles of #CO_2#.
Likewise, the #color(red)(1):color(blue)(3)# mole ratio that exists between benzene and water tells you that the reaction will produce 3 time more moles of water than the number of moles of benzene that reacted.
Since benzene is the limiting reagent all the avaialble moles will react. To determine exactly how many moles of benzene you're dealing with, use its molar mass
#105.5cancel("g") * ("1 mole"C_6H_6)/(78.11cancel("g")) = "1.351 moles"# #C_6H_6#
This means that the reaction will produce
#1.351cancel("moles"C_6H_6) * (color(green)(6)" moles"CO_2)/(color(red)(1)cancel("moles"C_6H_6)) = "8.106 moles"# #CO_2#
and
#1.351cancel("moles"C_6H_6) * (color(blue)(3)" moles"H_2O)/(color(red)(1)cancel("moles"C_6H_6)) = "4.053 moles"# #H_2O#
In order to get the pressure of the produced carbon dioxide gas at that temperature and in that volume, use the ideal gas law equation
#PV = nRT => P = (nRT)/V#
#P_(CO_2) = (8.106cancel("moles") * 0.082("atm" * cancel("L"))/(cancel("mol") * cancel("K")) * 300cancel("K"))/(540cancel("L"))#
#P_(CO_2) = "0.369 atm"#
I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the temperature of the gas
#P_(CO_2) = color(green)("0.37 atm")#
