To solve this problem, we'll start by using the ideal gas law and the formula for the root mean square (rms) velocity of a gas.

The ideal gas law is given by:

$$ PV = nRT $$

where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the universal gas constant, and $ T $ is the temperature. The rms velocity ($ v_{rms} $) of a gas is given by:

$$ v_{rms} = \sqrt{\frac{3RT}{M}} $$

where $ M $ is the molar mass of the gas. We need the rms velocities of gases $ X $ and $ Y $.

First, let's find the number of moles of each gas. For gas $ X $, let’s denote the molar mass as $ M_X $. The number of moles is:

$$ n_X = \frac{10 \, \text{g}}{M_X} $$

For the initial gas $ X $ at $ 300 \, \text{K} $ and $ 2 \, \text{atm} $, we can write:

$$ P_X V = n_X RT $$

When gas $ Y $ is added, the total pressure becomes $ 6 \, \text{atm} $. Let the molar mass of $ Y $ be $ M_Y $. The number of moles of gas $ Y $ added is:

$$ n_Y = \frac{80 \, \text{g}}{M_Y} $$

Now, the total number of moles in the vessel is $ n_X + n_Y $, and this total exerts a pressure of $ 6 \, \text{atm} $ at the same volume and temperature:

$$ P_{total} V = (n_X + n_Y) RT $$

Given the total pressure is $ 6 \, \text{atm} $, solving for the volume$ V $ and equating, we get two equations:

$$ 2V = \frac{10}{M_X} RT \quad ......\text{(i)} $$

$$ 6V = \left( \frac{10}{M_X} + \frac{80}{M_Y} \right) RT \quad ......\text{(ii)} $$

Dividing (ii) by (i) and solving for $ \frac{80}{M_Y} $, we get:

$$ \frac{80}{M_Y} = 2 \frac{10}{M_X} $$

$$ \frac{80}{M_Y} = \frac{20}{M_X} $$

$$ M_Y = 4 M_X $$

Now, applying the formula for the rms velocity, the ratio of rms velocities of $ X $ and $ Y $ is:

$$ \frac{v_{rms, X}}{v_{rms, Y}} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{4 M_X}{M_X}} = 2 $$

So, the ratio of root mean square velocities of $ X $ and $ Y $ is $ 2:1 $.

Thus, the correct answer is Option D: $ 2:1 $.