Answer
$+77.69\ kJ/mol$
Work Step by Step
Enthalpies of formation (kJ/mol):
$H_2O(l):-285.83, CO_2(g):393.509$
Enthalpy of combustion:
$-5156.1=4\times-285.83+10\times-393.509-x$
$x=+77.69\ kJ/mol$, which is the enthalpy of formation of naphthalene.
