A ferry boat has an internal volume of 1 m^3 and a weight of 50 kg...

Let $V$ be the total volume of the boat and $m$ be its mass.

When water starts coming in from the sides, the boat is completely submerged. The volume of water displaced is equal to the total volume of the boat, $V = 1 \text{ m}^3$.

The buoyant force acting on the boat is $F_B = V \rho_w g$, where $\rho_w = 1000 \text{ kg/m}^3$ is the density of water.

For the boat to just float, the total weight of the boat and the water inside it must equal the buoyant force. Let $V_{in}$ be the volume of water inside the boat.

$m g + V_{in} \rho_w g = V \rho_w g$

Substituting the given values:

$50 + V_{in} (1000) = 1 \times 1000$

$1000 V_{in} = 1000 - 50 = 950$

$V_{in} = \dfrac{950}{1000} = \dfrac{19}{20} \text{ m}^3$

The fraction of the boat's volume filled with water is $\dfrac{V_{in}}{V} = \dfrac{19/20}{1} = \dfrac{19}{20}$.