Answer
$\rho=1.05\times 10^4\frac{Kg}{m^3}$, silver
Work Step by Step
We know that
$\rho=\frac{m}{v}$
We plug in the known values to obtain:
$\rho=\frac{0.347}{(3.21cm(\frac{1m}{100cm}))^3}$
$\rho=1.05\times 10^4\frac{Kg}{m^3}$
Thus, the cube is silver as the density of silver is the same as calculated above.
