Molality ($m$) = (moles of solute) / (mass of solvent in kg). Molarity ($M$) = 3 mol/L. Mass of 1 L (1000 mL) of solution = density $\times$ volume = $1.25 \text{ g/mL} \times 1000 \text{ mL} = 1250 \text{ g}$. Mass of $NaCl$ (solute) in 1 L = moles $\times$ molar mass = $3 \text{ mol} \times 58.5 \text{ g/mol} = 175.5 \text{ g}$. Mass of solvent (water) = mass of solution - mass of solute = $1250 \text{ g} - 175.5 \text{ g} = 1074.5 \text{ g} = 1.0745 \text{ kg}$. Molality ($m$) = $3 \text{ mol} / 1.0745 \text{ kg} \approx 2.79 \text{ mol/kg}$ or $2.79 \text{ m}$. Option 2 and 3 are incorrect due to calculation errors, and Option 4 incorrectly assumes molarity equals molality when density is not $1 \text{ g/mL}$.