Given data

• Molarity of NaCl solution $= 3\,\mathrm{M}$
• Density of solution $= 1.0\,\mathrm{g\,mL^{-1}}$
• Molar mass of NaCl: $M_{\mathrm{NaCl}} = 23 + 35.5 = 58.5\,\mathrm{g\,mol^{-1}}$

Take 1 L of solution Since molarity is $3\,\mathrm{mol\,L^{-1}}$, in $1\,\mathrm{L}$ of solution: $\text{moles of NaCl} = 3\,\mathrm{mol}$

Mass of 1 L solution Density $= 1.0\,\mathrm{g\,mL^{-1}}$ $1\,\mathrm{L} = 1000\,\mathrm{mL}$ Hence mass of $1\,\mathrm{L}$ solution: $1000 \times 1.0 = 1000\,\mathrm{g}$

Mass of solute (NaCl) $\text{mass of NaCl} = 3 \times 58.5 = 175.5\,\mathrm{g}$

Mass of solvent $\text{mass of solvent} = \text{mass of solution} - \text{mass of solute}$ $= 1000 - 175.5 = 824.5\,\mathrm{g}$ $= 0.8245\,\mathrm{kg}$

Calculate molality Molality is moles of solute per kg of solvent: $m = \frac{3}{0.8245}$ $m \approx 3.6386\,\mathrm{m}$

Express in the required form We need: $m = \underline{\hspace{1cm}} \times 10^{-2}\,\mathrm{m}$ Since $3.6386 = 363.86 \times 10^{-2}$ Nearest integer $= 364$.

Final answer $\boxed{364}$