In case of H2O molecule, the enthalpy needed to break the two O - H bonds is not the same, i.e. 

H₂O (g) → H(g) + OH(g); ∆_a H₁^° = 502 kJ mol^-1

OH (g) → H(g) + O(g); ∆_a H₂^° = 427 kJ mol^-1. 

What should be the mean bond enthalpy of O - H bonds in case of H₂O molecule?

In case of ${\mathrm{H}}_2\mathrm{O}$ molecule, the enthalpy needed to break the two O-H bonds is not the same,
i.e.  ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)⟶\mathrm{H}\left(\mathrm{g}\right)+\mathrm{OH}\left(\mathrm{g}\right);{\mathrm{\Delta }}_{\mathrm{a}}{\mathrm{H}}_1^{\ominus }=502 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

$\mathrm{OH}\left(\mathrm{g}\right)⟶\mathrm{H}\left(\mathrm{g}\right)+\mathrm{O}\left(\mathrm{g}\right);{\mathrm{\Delta }}_{\mathrm{a}}{\mathrm{H}}_2^{\ominus }=427 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

The mean or average bond enthalpy is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule.
Average bond enthalpy = $\frac{502+427}{2}$ =  464.5kJ ${\mathrm{mol}}^{-1}$