Enthalpy of neutralization is given that is x.
And when 500ml of 2N HCl are mixed with 250 ml of 4 N NaOH,
then the mmoles required for the total neutalization for HCl is => N*V => 500*2 = 1000 mmoles
and for NaOH => 250*4 = 1000 mmoles.
Hence the complete neutralization would take place and therefore,
The enthalpy would remain same that is => x.