Equilibrium question Ethyl ethonate  with water?

This is a HOMOGENEOUS equilibrium (not a heterogeneous equilibrium!!!).  All the four components exist in one phase.  Hence, water should appear in the equilibrium law.

The mole ratio of the four components is 1 : 1 : 1 : 1.  Since equal amounts of ethanoic acid and ethanol were mixed initially, equal amounts of the ester and water would be formed at equilibrium.

                        CH₃CO₂C₂H₅ + H₂O ⇌ C₂H₅OH + CH₃CO₂H   Kc = 0.27
Eqm (mol dm⁻³):         y               y          0.42          0.42

At equilibrium:  Kc = [C₂H₅OH] [CH₃CO₂H] / ([CH₃CO₂C₂H₅][H₂O])
0.27 = 0.42² / y²
y = √(0.42² / 0.27)
y = 0.81

The answer:  C. 0.81 mol dm⁻³

Kc = [Ethanol][Ethanoic Acid] / [ Ethyl Ethanoate]

Water can be discounted !!!! 
[Ethyl Ethanoate] = [Ethanol][Ethanoic Acid] / Kc 
[ethyl ethanoate] = 0.42 mol dm^-3 *  0.42 mol dm^-3 / 0.27 
[ethyl ethanoate] = 0.65333.... ~ 0.65 mol dm^-3    ; Answer 'B'.