So we need #"(i) the volume of the room.........."#

And #"(ii) the number of air molecules.........."#.

#"(i) Volume"# #=# #13.5xx12.0xx10.0*ft^3=1620*ft^3#. (I never thought I would use #"feet and inches"#; I'll be bidding #"guineas"# at an auction next!)

#"Volume"# #=# #1620*ft^3xx28.2*L*ft^-3xx10^-3*m^3*L^-1=45.68*m^3#.

And now we need the molar volume at #20# #""^@C#, i.e. #293*K#

#V=(nRT)/(P)#

#=(1*molxx0.0821*L*atm*K^-1*mol^-1xx293*K)/(1*atm)=#

#=24.05*L#

And so if we divide the #"volume"# by the #"molar volume"#, and then multiply this number by the #"Avocado number"# (yes, I know but I am partial to avocadoes), we should reasonably get the number of air molecules in the room. It is a fact that #10^3L-=1*m^3#.

So #(45.68*m^3xx10^3*L*m^-3)/(24.05*L*mol^-1)~=1900*mol#

And thus #(ii)# there are #1900*molxx6.022xx10^23*mol^-1=1.14xx10^27# #"air molecules"#

Please check my calculations,

#"all care taken but no responsibility admitted"#.