To find the molality of the solution, we need to follow these steps:

1. Determine the number of moles of anhydrous $$\mathrm{CuSO}_4$$ in the solution.

The molarity (M) is given as $$2 \times 10^{-1}$$ M in a 500 mL solution. Hence, the number of moles of $$\mathrm{CuSO}_4$$ is:

$$ \text{Moles of } \mathrm{CuSO}_4 = \text{Molarity} \times \text{Volume in liters} = 2 \times 10^{-1} \times 0.5 = 0.1 \text{ moles} $$

2. Calculate the mass of the solution using the given density.

The density of the solution is given as $$1.25 \mathrm{~g/mL}$$. The volume of the solution is 500 mL. Therefore, the mass of the solution is:

$$ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 500 = 625 \text{ grams} $$

3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous $$\mathrm{CuSO}_4$$) from the total mass of the solution.

We know the number of moles of $$\mathrm{CuSO}_4$$, and we can find its molar mass.

$$ \text{Molar mass of } \mathrm{CuSO}_4 = 63.5 + 32 + 4 \times 16 = 159.5 \text{ g/mol} $$

Therefore, the mass of $$\mathrm{CuSO}_4$$ is:

$$ \text{Mass of } \mathrm{CuSO}_4 = \text{Number of moles} \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \text{ grams} $$

4. Calculate the mass of the solvent (water):

$$ \text{Mass of water} = \text{Mass of solution} - \text{Mass of } \mathrm{CuSO}_4 = 625 - 15.95 = 609.05 \text{ grams} $$

Convert the mass of water to kilograms:

$$ \text{Mass of water} = 609.05 \text{ grams} = 0.60905 \text{ kilograms} $$

5. Calculate the molality using the formula:

$$ \text{Molality} (\text{m}) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.1}{0.60905} = 0.164 \text{ mol/kg} $$

Converting to the desired unit:

$$ \text{Molality} (\text{m}) = 0.164 \times 10^3 = 164 \times 10^{-3} $$

Therefore, the molality of the solution is approximately 164 $$\times 10^{-3} \mathrm{~m}$$.