Step 1: Understanding the Concept:
According to Gauss's Law, the total electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space ($\varepsilon_0$).
A cube has six identical square faces. Due to the symmetrical placement of the charge at the center, the flux is distributed equally among all six faces.
Step 2: Key Formula or Approach:
Gauss's Law for total flux $\Phi_{\text{total}}$:
\[ \Phi_{\text{total}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \]
Flux through one face of a symmetrical cube:
\[ \Phi_{\text{one face}} = \frac{\Phi_{\text{total}}}{6} \]
Step 3: Detailed Explanation:
The charge placed at the center of the cube is $q_{\text{enclosed}} = Q$ (ignoring the micro prefix as the options only present $Q$, effectively treating $Q$ as the variable representing the charge magnitude).
The total flux passing through all six faces of the cube is:
\[ \Phi_{\text{total}} = \frac{Q}{\varepsilon_0} \]
Because the charge is exactly at the center, the electric field is symmetric with respect to all six faces. The flux passing through any single face is one-sixth of the total flux:
\[ \Phi_{\text{one face}} = \frac{1}{6} \left( \frac{Q}{\varepsilon_0} \right) = \frac{Q}{6\varepsilon_0} \]
The question asks for the flux through two opposite faces of the cube. The flux through two faces will simply be twice the flux through one face:
\[ \Phi_{\text{two faces}} = 2 \times \Phi_{\text{one face}} \]
\[ \Phi_{\text{two faces}} = 2 \times \left( \frac{Q}{6\varepsilon_0} \right) \]
\[ \Phi_{\text{two faces}} = \frac{Q}{3\varepsilon_0} \]
Step 4: Final Answer:
The flux through two opposite faces is $\frac{Q}{3\varepsilon_0}$.