Let volume of wooden block be V.

Volume of block submerged in water = 2\over 3V

Density of water = 1000 kg m^{-3}

By the principle of floatation,

{Volume\ of\ immersed\ part\over Total\ volume} = {Density\ of\ wood\over Density\ of\ water}⇒ {2V/3\over V} = {Density\ of\ wood\over Density\ of\ water}

∴ Density of wood = {2\over 3}×1000 = 667 kg m^{-3}

(b) Volume of block submerged in oil = {3\over 4}V

By the principle of floatation,

{Volume\ of\ immersed\ part\over Total\ volume} = {Density\ of\ wood\over Density\ of\ oil}⇒ {3V/4\over V} = {667\over Density\ of\ oil}

⇒ Density of oil = 4\over 3 × 667 = 889.33 kg m^{-3}