For this question we have to assume the density of fresh water to be 1g/cc (1g/mL) which is 1kg/L or 1000kg/m3.
Volume V of the cylindrical drum of radius r=30cm=0.3m is given by V=πr2h where h=1m, making V=0.09π m3, and the mass of oil of density 0.8(0.09π)(1000)kg=72πkg. Total mass of drum=72π+30 kg. This is equal to the mass of fresh water displaced, which can be converted to a volume, knowing the density of fresh water: mass/volume=1000, so volume v of water=mass/1000=(72π+30)/1000=0.072π+0.03 m3. Compare this with the volume of the drum: (0.072π+0.03)/0.09π=0.9061 or 90.61%. The reserve buoyancy is therefore about 9.39%.