Determining the Average Bond Enthalpy for the C-F bond
Your approach to use Hess's Law is reasonable!

\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]

That's the enthalpy for $\ce{CF4}$ - a molecule with four $\ce{C-F}$ bonds. 

The average $\ce{C-F}$ bond enthalpy is smaller:

\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx  427\ \mathrm{kJ\cdot mol^{-1}}\]