The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in ms$$^{-1}$$ is: [k$$_B$$ = 1.4 $$\times$$ 10$$^{-23}$$ J/K; m$$_{He}$$ = 7 $$\times$$ 10$$^{-27}$$ kg]

We are asked to find the “thermal velocity” of a helium atom at room temperature. In kinetic theory this quantity is usually taken as the root-mean-square speed, given by the well-known formula

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

Here $$k_B$$ is Boltzmann’s constant, $$T$$ is the absolute temperature and $$m$$ is the mass of one atom (or molecule) of the gas.

We now substitute the data provided in the question. We have

$$k_B = 1.4 \times 10^{-23}\;{\rm J\,K^{-1}}, \qquad T = 300\;{\rm K}, \qquad m_{He} = 7 \times 10^{-27}\;{\rm kg}.$$

First we calculate the numerator $$3 k_B T$$ step by step.

$$3 k_B T = 3 \times (1.4 \times 10^{-23}) \times 300.$$

Multiply the numbers in the brackets:

$$1.4 \times 300 = 420.$$

So

$$3 k_B T = 3 \times 420 \times 10^{-23}.$$

Now, $$3 \times 420 = 1260,$$ therefore

$$3 k_B T = 1260 \times 10^{-23}\;{\rm J}.$$

We rewrite the power of ten so that there is only one non-zero digit before the decimal point:

$$1260 \times 10^{-23} = 1.26 \times 10^{3} \times 10^{-23} = 1.26 \times 10^{-20}\;{\rm J}.$$

Next we form the ratio with the mass of a helium atom:

$$\dfrac{3 k_B T}{m_{He}} = \dfrac{1.26 \times 10^{-20}}{7 \times 10^{-27}}.$$

We divide the coefficients and subtract the exponents of ten:

$$\dfrac{1.26}{7} \approx 0.18, \qquad 10^{-20} \,/\, 10^{-27} = 10^{7}.$$

So

$$\dfrac{3 k_B T}{m_{He}} \approx 0.18 \times 10^{7} = 1.8 \times 10^{6}\;{\rm (m/s)^2}.$$

Finally, we take the square root to obtain the speed:

$$v_{\text{rms}} = \sqrt{1.8 \times 10^{6}}.$$

We separate the numerical and the power-of-ten parts:

$$\sqrt{1.8} \approx 1.34, \qquad \sqrt{10^{6}} = 10^{3}.$$

Therefore

$$v_{\text{rms}} \approx 1.34 \times 10^{3}\;{\rm m\,s^{-1}}.$$

This value is closest to $$1.3 \times 10^{3}\;{\rm m\,s^{-1}}.$$

Hence, the correct answer is Option D.