1. Given data

   • Mass of cube: $m = 400\text{ g} = 0.4\text{ kg}$
   • Edge of cube: $a = 10\text{ cm} = 0.1\text{ m}$
   • Density of water: $\rho_w = 1000\,\text{kg m}^{-3}$

2. Total volume of the cube $V_{\text{cube}} = a^3 = (10\text{ cm})^3 = 1000\text{ cm}^3$

3. Condition for floating For a floating body, weight of cube = buoyant force.

   So, mass of displaced water = mass of cube.

   Hence submerged volume is $V_{\text{sub}} = \frac{m}{\rho_w} = \frac{0.4}{1000}\,\text{m}^3 = 4\times 10^{-4}\,\text{m}^3$

4. Convert submerged volume into cm$^3$ Since $1\,\text{m}^3 = 10^6\,\text{cm}^3$, $V_{\text{sub}} = 4\times 10^{-4}\times 10^6 = 400\text{ cm}^3$

5. Volume outside water $V_{\text{outside}} = V_{\text{cube}} - V_{\text{sub}} = 1000 - 400 = 600\text{ cm}^3$

6. Option check

   • A: $400\text{ cm}^3$ ❌
   • B: $600\text{ cm}^3$ ✅
   • C: $1400\text{ cm}^3$ ❌
   • D: $4000\text{ cm}^3$ ❌

Therefore, the correct answer is B.