$$ \text{Molarity of NaCl} = 3 \text{ M} \quad (\text{3 moles in 1 L of solution}) $$

Given the density of the solution is $$1.25 \, \text{g/mL},$$ the total mass of 1 liter (1000 mL) of the solution is:

$$ 1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g} $$

To find the molality, we need the mass of the solvent (water). First, calculate the mass of NaCl in 1 L of solution. The molar mass of NaCl is approximately:

$$ 23 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} \approx 58.45 \, \text{g/mol} $$

Thus, the mass of NaCl is:

$$ 3 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 175.35 \, \text{g} $$

Now, the mass of the solvent (water) is:

$$ 1250 \, \text{g (solution)} - 175.35 \, \text{g (NaCl)} \approx 1074.65 \, \text{g} $$

Convert the mass of the solvent to kilograms:

$$ 1074.65 \, \text{g} \div 1000 \approx 1.07465 \, \text{kg} $$

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent:

$$ m = \frac{3 \, \text{mol}}{1.07465 \, \text{kg}} \approx 2.79 \, \text{mol/kg} $$

Thus, the molality of the solution is approximately $$2.79 \, \text{m}.$$