An infinitely long wire has uniform linear charge density $$\lambda = 2$$ nC/m. The net flux through a Gaussian cube of side length $$\sqrt{3}$$ cm, if the wire passes through two corners of the cube that are maximally displaced from each other, would be $$x$$ Nm$$^2$$C$$^{-1}$$, where x is :
[Neglect any edge effects and use $$\frac{1}{4\pi\epsilon_0}= 9\times10^{9}$$ SI units]

A long straight wire carries uniform linear charge density $$\lambda = 2\,\text{nC m}^{-1}=2\times10^{-9}\,\text{C m}^{-1}$$.
We have to find the electric-flux through a Gaussian cube of side length $$a=\sqrt{3}\,\text{cm}=0.01\sqrt{3}\,\text{m}$$ when the wire passes through the two diagonally opposite corners of the cube.

Step 1 : Length of the wire lying inside the cube
For a cube of side $$a$$, the space-diagonal length is $$a\sqrt{3}$$.
Because the straight wire enters at one corner and exits at the diametrically opposite corner, the portion of the wire enclosed by the cube equals the space-diagonal.
Hence
$$ \ell_{\text{in}} = a\sqrt{3} $$

Substituting $$a=\sqrt{3}\,\text{cm}$$,
$$ \ell_{\text{in}} = (\sqrt{3}\,\text{cm})\sqrt{3}=3\,\text{cm}=0.03\,\text{m} $$

Step 2 : Charge enclosed by the Gaussian surface
Using $$q_{\text{enc}} = \lambda \ell_{\text{in}}$$,
$$ q_{\text{enc}} = (2\times10^{-9})\times(0.03)=6\times10^{-11}\,\text{C} $$

Step 3 : Electric-flux through the cube
Gauss’s law states $$\Phi = \dfrac{q_{\text{enc}}}{\varepsilon_0}$$ $$-(1)$$
With $$\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$$, substitute in $$(1)$$:
$$ \Phi = \dfrac{6\times10^{-11}}{8.854\times10^{-12}} = 6.784\times10^{0}\;\text{N m}^2\text{C}^{-1} $$

Step 4 : Expressing the result in terms of $$\pi$$
Calculate $$2.16\pi$$:
$$ 2.16\pi = 2.16 \times 3.1416 \approx 6.79 $$ which matches the obtained value (6.784). Hence
$$ \Phi = 2.16\pi\; \text{N m}^2\text{C}^{-1} $$

Therefore, $$x = 2.16\pi$$ and the correct option is Option D.