Why is water taken in calculating K, equilibrium constant
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.