Answer
See answer below.
Work Step by Step
Ethanol:
Mass of $CO_2$: $6/1\ mol\times44.01\ g/mol=264.06\ g$
Volume: $1\ mol\times 94.11\ g/mol\div785\ g/L=0.120\ L$
$264.06\ g\div 0.120\ L= 2202.6\ g/L$
Octane:
Mass of $CO_2$: $8/1\ mol\times44.01\ g/mol=352.08\ g$
Volume: $1\ mol\times 114.23\ g/mol\div699\ g/L=0.163\ L$
$352.08\ g\div 0.163\ L= 2154.4\ g/L$