Two point charges \(2 q\) and \(q\) are placed at vertex \(A\) and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :

Explanation:
To find the total electric flux passing through the cube, we apply Gauss's Law and the Principle of Superposition. We will calculate the flux contribution of each charge individually and sum them up.

1. Contribution of charge \( 2q \) at vertex A:
A charge placed at the corner (vertex) of a cube is shared equally among 8 such adjacent cubes. Therefore, only \( \frac{1}{8} \) of the charge is effectively "inside" our cube.
Flux due to the vertex charge, \( \phi_1 = \frac{1}{8} \left( \frac{2q}{\epsilon_0} \right) = \frac{q}{4\epsilon_0} \)

2. Contribution of charge \( q \) at the centre of face CDEF:
A charge placed at the centre of a face is shared equally between 2 adjacent cubes. Therefore, only \( \frac{1}{2} \) of the charge is effectively "inside" our cube.
Flux due to the face-centred charge, \( \phi_2 = \frac{1}{2} \left( \frac{q}{\epsilon_0} \right) = \frac{q}{2\epsilon_0} \)

3. Total electric flux (\( \phi \)) through the cube:
\( \phi = \phi_1 + \phi_2 \)
\( \phi = \frac{q}{4\epsilon_0} + \frac{q}{2\epsilon_0} \)
\( \phi = \frac{q + 2q}{4\epsilon_0} = \frac{3q}{4\epsilon_0} \)

Hence, Option D is the correct choice.