I can't find reaction enthalpy ∆H for water formation reaction
Here is the tally of bonds:


One $\ce{H-H}$ bond broken
One half $\ce{O=O}$ double bond broken
Two $\ce{O-H}$ bonds made


Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $\ce{O-H}$ bond depends on what else is attached to the oxygen. For H-H and O=O, this is no problem because there is only one molecule, respectively, with that kind of bond.

$$\Delta H = \pu{436 kJ/mol} + \pu{249.5 kJ/mol} - \pu{928 kJ/mol} =  \pu{-242.5 kJ/mol}$$

We got the right sign and are not too far off in magnitude.

Why is the physical state important?

In solids and liquids, there are intermolecular interactions that contribute to the enthalpy. This is why the enthalpy of sublimation and boiling is positive - it takes energy to break the intermolecular forces. Here are the values for enthalpy of formation for water in the two states at room temperature:

$$\Delta H_\mathrm{f}(\ce{H2O(l)}) = \pu{-285.8 kJ/mol}$$

$$\Delta H_\mathrm{f}(\ce{H2O(g)}) = \pu{-241.8 kJ/mol}$$

And here is the enthalpy of vaporization

$$\Delta H_\mathrm{vap}(\ce{H2O(g)}) = \pu{43.99 kJ/mol}$$