To solve this problem, we need to determine the new pressure of the moist gas when the volume of the container is doubled while maintaining the same temperature. First, we need to separate the contributions of the gas and the water vapor to the initial pressure and then apply Boyle's Law for the gas component only. The steps are as follows:

1. Initial pressure of the moist gas (gas + water vapor) is given as $$4 \, \text{atm}$$.

The vapour pressure of water at $$27^{\circ} \mathrm{C}$$ is $$0.4 \, \text{atm}$$.

2. The partial pressure of the dry gas component can be calculated by subtracting the vapor pressure of water from the initial pressure:

$$ P_{\text{dry gas (initial)}} = 4 \, \text{atm} - 0.4 \, \text{atm} = 3.6 \, \text{atm} $$

3. When the volume of the container is doubled, the pressure of the dry gas component will change according to Boyle's Law, which states:

$$ P_1 V_1 = P_2 V_2 $$

Since the temperature remains constant and the volume is doubled (i.e., $$V_2 = 2V_1$$), we can find the new pressure of the dry gas ($$P_{\text{dry gas (new)}}$$) with the equation:

$$ 3.6 \, \text{atm} \times V_1 = P_{\text{dry gas (new)}} \times 2V_1 $$

Solving for $$P_{\text{dry gas (new)}}$$:

$$ P_{\text{dry gas (new)}} = \frac{3.6 \, \text{atm}}{2} = 1.8 \, \text{atm} $$

4. The vapor pressure of water remains unchanged at $$0.4 \, \text{atm}$$ because it depends only on the temperature, not on the volume.

5. The new total pressure of the moist gas is the sum of the new pressure of the dry gas and the unchanged vapor pressure of water:

$$ P_{\text{moist gas (new)}} = P_{\text{dry gas (new)}} + P_{\text{water vapor}} = 1.8 \, \text{atm} + 0.4 \, \text{atm} = 2.2 \, \text{atm} $$

6. Finally, as requested, we express the new pressure in the form of $$\times 10^{-1} \, \text{atm}$$ and find the nearest integer:

$$ 2.2 \, \text{atm} = 22 \times 10^{-1} \, \text{atm} $$

Therefore, the nearest integer is 22.

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