1. **State the problem:** A particle with charge $q = 1.8$ microcoulombs ($1.8 \times 10^{-6}$ C) is placed at the center of a Gaussian cube with edge length $L = 55$ cm ($0.55$ m). We need to find the net electric flux $\Phi_E$ through the surface of the cube. 2. **Relevant formula:** According to Gauss's law, the net electric flux through a closed surface is given by: $$\Phi_E = \frac{q_{\text{enc}}}{\epsilon_0}$$ where $q_{\text{enc}}$ is the total charge enclosed by the surface and $\epsilon_0$ is the permittivity of free space, $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2$. 3. **Apply the formula:** The charge enclosed by the Gaussian surface is $q = 1.8 \times 10^{-6}$ C. 4. **Calculate the flux:** $$\Phi_E = \frac{1.8 \times 10^{-6}}{8.854 \times 10^{-12}}$$ 5. **Perform the division:** $$\Phi_E = 2.033 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C}$$ 6. **Interpretation:** The net electric flux through the cube's surface is approximately $2.03 \times 10^{5}$ N·m$^2$/C. This result is independent of the cube's size or shape, as Gauss's law depends only on the enclosed charge.