This is a problem involving the concept of buoyancy and equilibrium. Let's solve it step-by-step.

Problem Breakdown

• A block of ice is floating in the sea, and a penguin with a mass of $20 \, \text{kg}$ stands on it.
• The block submerges just to the level of the water, meaning it’s in equilibrium (not sinking further or rising).
• Given:
  • Mass of the penguin: $20 \, \text{kg}$
  • Density of ice: $0.9 \, \text{kg/L}$

We need to determine the mass of the ice block.

Step-by-Step Solution

1. Buoyant Force and Equilibrium Condition:
   For the block to be in equilibrium, the combined weight of the ice block and the penguin must equal the buoyant force exerted by the water on the block.

   The buoyant force ($F_b$) is given by: $F_b = \text{(Weight of the displaced water)}$

2. Weight of the Block + Penguin:
   The total weight acting on the system (block + penguin) is: $W_{\text{total}} = (M_{\text{ice}} + M_{\text{penguin}}) \times g$ where $M_{\text{ice}}$ is the mass of the ice block, $M_{\text{penguin}} = 20 \, \text{kg}$, and $g$ is the acceleration due to gravity (assumed to be $9.81 \, \text{m/s}^2$, though it will cancel out in the equations).

3. Buoyant Force of Water Displaced:
   Since the block is fully submerged to the water level, the volume of the ice block equals the volume of the displaced water.

   Using the density of ice, we can express the volume of the ice block $V_{\text{ice}}$ as: $V_{\text{ice}} = \frac{M_{\text{ice}}}{\rho_{\text{ice}}}$ where $\rho_{\text{ice}} = 0.9 \, \text{kg/L}$.

   The mass of the displaced water (and hence the buoyant force) is: $M_{\text{water}} = V_{\text{ice}} \times \rho_{\text{water}}$ where $\rho_{\text{water}}$ is the density of seawater (approximately $1 \, \text{kg/L}$ for this problem).

4. Setting up the Equation for Equilibrium:
   Since $M_{\text{water}} = M_{\text{ice}} + M_{\text{penguin}}$, we have: $V_{\text{ice}} \times \rho_{\text{water}} = M_{\text{ice}} + 20$

   Substituting (