Apiece of metal weighs 300 n in air and 232.5 n when submerged in water. find the density of the metal.

Given:
F1 = 300n
F2 = 232.5n
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If we neglect F Archimedes in air we will have the following:

F1 = mg
F2 in water will be [mg - (F(A) in water)]
IF we say Density of metal is: R(m) and density of water is R(w)
The volume of metal is -> v
m=R(m) * v    =>   F = R(m) * v * g
On the other side we know that 
F(A) = R(w) * v * g
now we can write the following

F2 = mg - F(A) = R(m)*v*g - R(w)*v*g = v*g*[R(m) - R(w)]
If we divide F1 by F2 we will have:
F1/F2 = [R(m)*v*g] / [v*g*(R(m)-R(w))] = R(m)/(R(m)-R(w))
Now we can write the values of F1 and F2 and we will get:


From here we can simplify this equation and will get the answer
If nowhere it's written about R(w), we should say that R(w) ≈ 1000 kg/m^3
300/232.5 = R(m)/(R(m) - 1000)
300*R(m) - 300*1000 = 232.5*R(m)
300*R(m) - 232.5*R(m) = 300,000
67.5R(m) = 300,000
R(m) = 300,000/67.5 ≈ 4,444.4 kg/m^3

Sorry i don’t know what to do

Domain eukarya; kingdom plantae should be the answer.