The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:

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

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

In this case, we are given that the initial temperature is $$200 \, K$$ and the final temperature is $$800 \, K$$. Let the r.m.s speed at $$200 \, K$$ be $$v_0$$, then:

$$v_0 = \sqrt{\frac{3R \cdot 200}{M}}$$

Now, we want to find the r.m.s speed at $$800 \, K$$, let's call this $$v_1$$:

$$v_1 = \sqrt{\frac{3R \cdot 800}{M}}$$

Now, divide $$v_1$$ by $$v_0$$:

$$\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}$$

Simplify the expression:

$$\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$$

So, $$v_1 = 2v_0$$.

Hence, the r.m.s speed of the gas at $$800 \, K$$ will be 2 times the r.m.s speed of the gas at $$200 \, K$$.