1. **State the problem:** A particle with charge $1.8$ microcoulombs is placed at the center of a Gaussian cube with edge length $55$ cm. We need to find the net electric flux through the surface of the cube. 2. **Relevant formula:** The net electric flux $\Phi$ through a closed surface enclosing a charge $q$ is given by Gauss's law: $$\Phi = \frac{q}{\epsilon_0}$$ where $\epsilon_0$ is the permittivity of free space, $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2$. 3. **Important notes:** - The shape and size of the Gaussian surface do not affect the net flux, only the enclosed charge matters. - The charge must be in coulombs for the formula. 4. **Convert charge to coulombs:** $$q = 1.8 \text{ microcoulombs} = 1.8 \times 10^{-6} \text{ C}$$ 5. **Calculate the net electric flux:** $$\Phi = \frac{1.8 \times 10^{-6}}{8.854 \times 10^{-12}}$$ 6. **Perform the division:** $$\Phi = 2.033 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C}$$ **Final answer:** The net electric flux through the surface of the cube is approximately $$\boxed{2.03 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C}}$$