Answer
See answer below.
Work Step by Step
Enthalpies of formation (kJ/mol):
$C_2H_5OH(l):-277.0, CO_2(g): -393.509, H_2O(g): -241.83$
Enthalpy of combustion:
$2\times-393.509+3\times-241.83--277.0=-1235.5\ kJ/mol$
In 100g:
$100\ g\div 46.07\ g/mol\times-1235.5\ kJ/mol=-2682\ kJ$