The temperature of an ideal gas in three dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K is:
$$m_e$$ = mass of electron = $$9 \times 10^{-31}$$ kg, $$h$$ = Planck constant = $$6.6 \times 10^{-34}$$ J s, $$k_B$$ = Boltzmann constant = $$1.38 \times 10^{-23}$$ J K$$^{-1}$$

We are asked to find the de-Broglie wavelength of an electron present in an ideal gas whose temperature is $$T = 300 \text{ K}$$.

The average kinetic energy per molecule of an ideal gas in three dimensions is given by the well-known equipartition theorem:

$$\frac{3}{2}k_B T = \frac{1}{2}m v_{\text{rms}}^{\,2}.$$

Here $$k_B$$ is the Boltzmann constant, $$m$$ is the mass of the particle and $$v_{\text{rms}}$$ is the root-mean-square speed. We first solve this equation for $$v_{\text{rms}}$$.

Multiplying both sides by 2, we get

$$3 k_B T = m v_{\text{rms}}^{\,2}.$$

Dividing by $$m$$ and then taking the square root,

$$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}.$$

Now we substitute the given numerical values: $$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1},\; T = 300\ \text{K},\; m = 9 \times 10^{-31}\ \text{kg}.$$

First the product $$k_B T$$:

$$k_B T = (1.38 \times 10^{-23})(300) = 1.38 \times 300 \times 10^{-23} = 414 \times 10^{-23} = 4.14 \times 10^{-21}\,\text{J}.$$

Then the factor $$3 k_B T$$ appearing in the numerator:

$$3 k_B T = 3 \times 4.14 \times 10^{-21} = 12.42 \times 10^{-21} = 1.242 \times 10^{-20}\,\text{J}.$$

Now we divide by the mass of the electron:

$$\frac{3 k_B T}{m} = \frac{1.242 \times 10^{-20}}{9 \times 10^{-31}} = \frac{1.242}{9} \times 10^{-20+31} = 0.138 \times 10^{11} = 1.38 \times 10^{10}\ \text{(m}^2\text{/s}^2).$$

Taking the square root gives the rms speed:

$$v_{\text{rms}} = \sqrt{1.38 \times 10^{10}} = \sqrt{1.38}\,\sqrt{10^{10}} \approx 1.175 \times 10^{5}\ \text{m s}^{-1}.$$

With the speed known, we use the de-Broglie relation

$$\lambda = \frac{h}{m v}$$

and for thermal motion we put $$v = v_{\text{rms}}$$. Stating the values, $$h = 6.6 \times 10^{-34}\ \text{J s}$$, $$m = 9 \times 10^{-31}\ \text{kg}$$, and $$v_{\text{rms}} \approx 1.175 \times 10^{5}\ \text{m s}^{-1}$$.

The denominator $$m v_{\text{rms}}$$ is

$$m v_{\text{rms}} = (9 \times 10^{-31})(1.175 \times 10^{5}) = 9 \times 1.175 \times 10^{-31+5} = 10.575 \times 10^{-26} = 1.0575 \times 10^{-25}\ \text{kg m s}^{-1}.$$

Hence the wavelength is

$$\lambda = \frac{6.6 \times 10^{-34}} {1.0575 \times 10^{-25}} = \frac{6.6}{1.0575} \times 10^{-34+25} \approx 6.24 \times 10^{-9}\ \text{m}.$$

Since $$1\ \text{nm} = 10^{-9}\ \text{m}$$, we convert the answer:

$$\lambda \approx 6.24\ \text{nm}.$$

The option that is nearest to this value is 6.26 nm.

Hence, the correct answer is Option D.