Answer
See answer below.
Work Step by Step
Enthalpies of formation (kJ/mol):
$CO(g):-110.525, CO_2(g): -393.509, H_2O(g): -241.83$
Enthalpy of combustion:
$-393.509\ kJ/mol$
Enthalpy of water gas reaction:
$-110.525--241.83=+131.31\ kJ/mol$
Number of moles of carbon:
$1000\ g\div 12.011\ g/mol=83.25\ mol$
Heat required:
$83.25\ mol\times 131.31\ kJ/mol=10.93\ MJ$
Mass of carbon for combustion:
$10932\ kJ\div(393.509\ kJ/mol\div12.011\ g/mol)=333.7\ g= 0.33\ kg$