1 mol ethanoic acid + 2 mol ethanol -> equilibrium
diluted to 250cm3 and 25cm3 is taken out
0.0155 mol of NaOH required to neutralise the 25cm3 which means there are 0.0155 mol of acid in the 25cm3 and therefore 0.155 mol of acid in 250cm3 in equilibrium.
As the acid dropped from 1 to 0.155 mol, it dropped by 0.845, and so did the ethanol (from 2 to 1.155).
the water and the ethyl ethanoate increased by the same amount and are now both 0.845 each.
putting it into the kc calculation, it is products over reactants which is (0.845*0.845)/(0.155*1.155) = 3.9884