It is more like a physics, Archimedes principle, or density problem.
The volume of lead is 0.25 m^3, for which density is 11,000 kg/m^3.
The mass is therefore ρ v = 11000*0.25 = 2750 kg.
Neglecting the weight of the rest of the boat, and using given density of sea-water as 1000 kg/m^3 (in fact, it is around 1025 kg/m^3), the total buoyant volume required is
volume = (2750 kg) / (1000 kg/m^3) = 2.75 m^3
