The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is $$v_0$$. Then, r.m.s. speed of the gas at 800 K will be:

When the temperature of an ideal gas increases from $$200$$ K to $$800$$ K, the r.m.s. speed follows the relation $$v_{rms} = \sqrt{\dfrac{3kT}{m}}$$, implying that $$v_{rms}\propto\sqrt{T}$$.

Accordingly, the ratio of r.m.s. speeds at $$800$$ K and $$200$$ K becomes

$$\dfrac{v_{rms}(800)}{v_{rms}(200)} = \sqrt{\dfrac{800}{200}} = \sqrt{4} = 2\,. $$

Since the r.m.s. speed at $$200$$ K is given as $$v_0$$, the speed at $$800$$ K is

$$v_{rms}(800) = 2 \times v_0 = 2v_0\,. $$

Therefore, the r.m.s. speed at $$800$$ K is \boxed{2v_0}.