Since the density of gasoline is not given, let us suppose it is 750 kg/m3 and the density of steel is 7800 kg/m3.
There are 2 constraints: 1. volume of steel + volume of gasoline = volume of water and at the threshold of sinking, 2. mass of steel + mass of gasoline = mass of water
240L * 1m³ / 1000L = 0.240 m³
first: Vs + Vg = Vw = Vs + 0.240m³
second: ρs*Vs + ρg*Vg = ρw*Vw → substitute for Vw, Vg and all the densities
7800*Vs + 750*0.240 = 1000*(Vs + 0.240) (7800 - 1000)*Vs = 240 - 750*0.240 Vs = 60 / 6800 = 0.00882 m³ = 8.82 L ≈ 9 L