Two conducting plates X and Y , each having a large surface area A...

Consider the surfaces of the plates as 1, 2 (for plate X) and 3, 4 (for plate Y) from left to right.

The total charge on the system of plates is $Q_{net} = Q + 0 = Q$.

The charge on the outermost surfaces (1 and 4) is equal to half of the total charge:

$q_1 = q_4 = \dfrac{Q_{net}}{2} = \dfrac{Q}{2}$

The charges on the inner surfaces (2 and 3) are equal and opposite:

$q_2 = Q - q_1 = \dfrac{Q}{2}$

$q_3 = 0 - q_4 = -\dfrac{Q}{2}$

The electric field at a point to the right of the plates is the vector sum of the fields due to all four surfaces. Taking the right direction as positive:

$E = \dfrac{q_1}{2A\varepsilon_0} + \dfrac{q_2}{2A\varepsilon_0} + \dfrac{q_3}{2A\varepsilon_0} + \dfrac{q_4}{2A\varepsilon_0}$

$E = \dfrac{1}{2A\varepsilon_0} \left( \dfrac{Q}{2} + \dfrac{Q}{2} - \dfrac{Q}{2} + \dfrac{Q}{2} \right)$

$E = \dfrac{Q}{2A\varepsilon_0}$

Since the result is positive, the electric field is directed towards the right.