To find the net flux through a Gaussian cube when an infinitely long wire with a uniform linear charge density $ \lambda = 2 \text{ nC/m} $ passes through the cube's diagonal corners, we use Gauss's Law. The relevant expression is:

$ \phi = \frac{q_{\text{enc}}}{\varepsilon_0} $

Here, $ q_{\text{enc}} $ is the charge enclosed by the Gaussian surface, and $ \varepsilon_0 $ is the permittivity of free space.

Given:

The wire passes through two opposite corners of the cube.

The side length of the cube, $ a = \sqrt{3} \text{ cm} = \sqrt{3} \times 10^{-2} \text{ m} $.

The length of the wire inside the cube is equal to its diagonal, which is $ \sqrt{3} \times a $.

Thus, the charge enclosed $ q_{\text{enc}} $ is:

$ q_{\text{enc}} = \lambda \times \text{(length of wire inside the cube)} = \lambda \times \sqrt{3} \times a $

Substitute the given values:

$ \lambda = 2 \times 10^{-9} \text{ C/m}, \quad a = \sqrt{3} \times 10^{-2} \text{ m} $

Calculating:

$ q_{\text{enc}} = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} $

Now apply Gauss's Law for net flux:

$ \phi = \frac{\lambda \cdot \sqrt{3} \cdot a}{\varepsilon_0} $

Substitute $\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}$:

$ \phi = 2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} \times 36 \pi \times 10^9 $

$ \phi = 2.16 \pi \text{ Nm}^2\text{C}^{-1} $

Thus, the net flux through the Gaussian cube is $ 2.16 \pi \text{ Nm}^2\text{C}^{-1} $.