Why is water taken in calculating K, equilibrium constant For the given question,

I was trying to solve it but I was not considering the concentration of water in the calculation (because water is a pure liquid and its concentration is supposed to be taken as 1), but then I needed the volume to be able to calculate the answer. I saw the solution,

Is it taken here because the reactants are organic and thus water would probably not be the solvent?
The question they gave you is somewhat flawed, in that there are four compounds and you don't know the range of their mutual solubility.  

To really answer the question, you would need information such as from Solubility, liquid–liquid equilibrium and critical states for the quaternary system acetic acid–ethanol–ethyl acetate–water at 293.15 K Fluid Phase Equilibria vol. 313, pages 46–51. 

The reactants, ethanol and acetic acid, are completely miscible; however, ethyl acetate only has limited solubility in water. 

When you evaluate the equilibrium constant, all the reacting compounds, including water, which are in a single phase, must be included.