A sphere held at the surface of a column of water is released to fall through the water. List the forces that act on the sphere.

Side length of the cube, $L = 0.3 \text{ cm} = 0.003 \text{ m}$.
Density of wood, $\rho_{\text{wood}} = 700 \text{ kg m}^{-3}$.
Density of water, $\rho_{\text{water}} = 1000 \text{ kg m}^{-3}$.
When an object floats, the weight of the object equals the upthrust force.
Weight of cube = Upthrust force

$$m_{\text{cube}}g = m_{\text{displaced water}}g$$ $$\rho_{\text{wood}} V_{\text{cube}} = \rho_{\text{water}} V_{\text{submerged}}$$

Let $h$ be the height of the cube submerged in water.
The volume of the cube is $V_{\text{cube}} = L^3$.
The volume of water displaced (submerged volume) is $V_{\text{submerged}} = L^2 h$.
Substitute these into the flotation equation:

$$\rho_{\text{wood}} L^3 = \rho_{\text{water}} L^2 h$$

Divide both sides by $L^2$:

$$\rho_{\text{wood}} L = \rho_{\text{water}} h$$

Solve for $h$:

$$h = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} L$$

Substitute the given values:

$$h = \frac{700 \text{ kg m}^{-3}}{1000 \text{ kg m}^{-3}} \times 0.003 \text{ m}$$ $$h = 0.7 \times 0.003 \text{ m}$$ $$h = 0.0021 \text{ m}$$

This is the portion of the cube that is above the water surface.
Distance above surface $= L - h$

$$d_{\text{above}} = 0.003 \text{ m} - 0.0021 \text{ m}$$ $$d_{\text{above}} = 0.0009 \text{ m}$$

Convert the distance back to centimeters:

$$d_{\text{above}} = 0.0009 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}}$$ $$d_{\text{above}} = 0.09 \text{ cm}$$

The distance from the top of the cube to the water surface is $\boxed{\text{0.09 cm}}$.