Let the density of lump of plastic be D.The relative density of an object is given as-relative density of object=density of object/density of waterHere, relative density of object=0.4 density of water=1g/cm3By using the above values we get0.4=D/1So, D=0.4Now let us suppose that the total volume of the lump is V,volume inside the water is V1and the volume outside water is V2 Density of lump=0.4g/cm^3But, Density of lump=Mass of lump/total volume of lumpSo, we get 0.4=10g/VTherefore, total volume of lump,V=25 cm^3According to the archimede's principle, when a body is partially or fully immersed in a liquid, the bouyant force acting on the body is eaqual to the weight of the liquid displaced by it. As the relative density of lump is less than 1 so,the lump will float in water and hence, the bouyant force will be equal to the weight of the lump and by archimede's principle the weight of the liquid dispaced by it will be 10g.Now, the weight of water,that is,displaced is 10g and the density of the water is 1g/cm^3But density of this water=weight of water/volume of the waterHere, the volume of water dispaced must be equal to the volume of the lump inside the water,i.e.,V1So we get 1(density)=10g(weight)/V1Therefore, V1=10cm^3But we need to find out the volume outside the water,V2Therefore, V2=V-V1So V2=15cm^3