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]

Since the wire passes through two corners of the cube that are maximally separated, it passes along the body diagonal of the cube.

The side of the cube is  $$a=\sqrt{3}\,cm$$

Therefore, the length of the wire inside the cube is the body diagonal:  $$l=a\sqrt{3}$$

$$l=\sqrt{3}\times\sqrt{3}\,cm=3\,cm$$

$$l=3\times10^{-2}\,m$$

Hence, the charge enclosed by the cube is  $$q=\lambda l$$

Given,  $$\lambda=2\times10^{-9}\,C\,m^{-1}$$

Therefore,  $$q=(2\times10^{-9})(3\times10^{-2})$$

$$q=6\times10^{-11}\,C$$

By Gauss's law,  $$\Phi=\frac{q}{\epsilon_0}$$

Using  $$\frac{1}{4\pi\epsilon_0}=9\times10^9$$

we get  $$\frac{1}{\epsilon_0}=4\pi(9\times10^9)$$

Therefore,  $$\Phi=6\times10^{-11}\times4\pi\times9\times10^9$$

$$\Phi=2.16\pi\,Nm^2C^{-1}$$

Hence,  $${x=2.16\pi}$$

Hence, the correct option is D.