The Gaussian surface for the charge placed at the center of the cube will spread out equally through all sides of the cube. According to Gauss's Law, electric flux $\Phi$ through a closed surface is equal to the charge enclosed $Q$ divided by the permittivity of free space $\epsilon_0$:

$$\Phi_{\text{total}} = \frac{Q}{\epsilon_0}$$

Given that the cube has 6 faces, and due to symmetry, the flux through each face will be equal, the flux through any one face $\Phi_{\text{face}}$ is:

$$\Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{6\epsilon_0}$$

Since the charge is given in microcoulombs ($\mu\mathrm{C}$), we need to convert it to coulombs by multiplying with $10^{-6}$:

$$\Phi_{\text{face}} = \frac{Q \times 10^{-6}}{6\epsilon_0}$$

This matches option D, which means the correct flux through one face of the cube given a charge Q microcoulombs at the center is $$\frac{Q}{6\epsilon_0} \times 10^{-6}$$.