According to Archimedes' principle, for an object to float, the weight of the object must be equal to the weight of the fluid it displaces.
Let \(V_{total} = 1\) litre = 1000 cm\(^3\) be the total outer volume of the block.
Let \(V_{wood}\) be the volume of the wood material and \(V_{cavity}\) be the volume of the cavity.
Then \( V_{total} = V_{wood} + V_{cavity} \).
The volume of water displaced is half the total volume: \( V_{displaced} = \frac{1}{2} V_{total} = \frac{1}{2}(1000) = 500 \) cm\(^3\).
The weight of the displaced water is \( W_{water} = V_{displaced} \times \rho_{water} \times g = 500 \times 1 \times g \). (Density of water \(\rho_{water}=1\) g/cm\(^3\)).
The weight of the wooden block is the weight of the wood material only (the cavity is empty and has no weight).
Weight of the block \( W_{block} = \text{mass}_{wood} \times g = (V_{wood} \times \rho_{wood}) \times g \).
The specific gravity of wood is given as 3/4, so \( \rho_{wood} = \frac{3}{4} \rho_{water} = \frac{3}{4} \) g/cm\(^3\).
\( W_{block} = (V_{wood} \times \frac{3}{4}) \times g \).
Equating the weight of the block and the weight of the displaced water:
\( V_{wood} \times \frac{3}{4} \times g = 500 \times 1 \times g \).
\( V_{wood} = 500 \times \frac{4}{3} = \frac{2000}{3} \) cm\(^3\).
Finally, find the volume of the cavity.
\( V_{cavity} = V_{total} - V_{wood} = 1000 - \frac{2000}{3} = \frac{3000-2000}{3} = \frac{1000}{3} \) cm\(^3\).
\( V_{cavity} \approx 333.3 \) cm\(^3\) or 333.3 ml.