The entropy change involved in the isothermal reversible expansion of $$2$$ moles of an ideal gas from a volume of $$10 \, \text{dm}^3$$ to a volume of $$100 \, \text{dm}^3$$ at $$27^\circ \text{C}$$ is:

For an isothermal, reversible expansion of an ideal gas the entropy change is given by
$$\Delta S = nR \ln \frac{V_2}{V_1}$$
where

$$n = 2 \;\text{mol}, \qquad R = 8.314 \;\text{J mol}^{-1}\text{K}^{-1}, \qquad V_1 = 10 \,\text{dm}^3, \qquad V_2 = 100 \,\text{dm}^3$$

1. Evaluate the volume ratio:
$$\frac{V_2}{V_1} = \frac{100}{10} = 10$$

2. Substitute in the entropy formula:
$$\Delta S = 2 \times 8.314 \times \ln 10$$

3. Use $$\ln 10 = 2.302585$$ (keep four significant figures for accuracy).
$$\Delta S = 16.628 \times 2.302585 = 38.27 \;\text{J K}^{-1}$$

4. Rounding to three significant figures gives
$$\Delta S \approx 38.3 \;\text{J K}^{-1}$$

Because the calculation already includes the given 2 moles, this is the required entropy change for the process. The options show units of $$\text{J mol}^{-1}\text{K}^{-1}$$, but numerically Option D matches our result.

Hence, the correct choice is:
Option D which is: $$38.3 \, \text{J mol}^{-1} \, \text{K}^{-1}$$