How can there be an enthalpy of formation for gaseous water?
The only explanation that comes into my mind is that, 

$$\ce{H2 (g) + \frac{1}{2} O2 (g) -&gt; H2O (g) &lt;-&gt; H2O(l)}$$
 And 
$$\Delta H_f^\circ(\ce{H2O (g)}) = \Delta H_f^\circ (\ce{H2O (l)}) + \Delta H_\mathrm{vap} (\pu{25^\circ C}) = -285.8 + 44.99 = \pu{-241.8 kJ/mol}$$ almost as tabulated depending the sources, here the calculation is perform with tabulated values found in the 87th Edition of the Handbook of Chemistry and Physics. 

This is a hypothetical state but actually, vapour water is in equilibrium with liquid water even under the standard state. It's just that this equilibrium is very unlikely to go in direction of formation of the gas water. But if you find an ingenious way to vaporize water under these conditions then the heat of formation of gaseous water will be the same as I just calculated.



Also in the 87th Edition of the Handbook of Chemistry and Physics, it is said (5-4):


  The standard state of a pure gaseous substance is that of
  the substance as a (hypothetical) ideal gas at the standard
  state pressure.


Although I've never figured that out and I can't really tell you about the need for this value, it exists. Probably you'd like to have this kind of values when performing a calculation considering a particular state but as far as I used to do, I've never used this one.