Given
Wair = 297 N
Wwater = 263 N
Density of water = 1000 kg/m^3
Acceleration due to gravity = 9.8
Procedure
We can calculate the thrust of the water with the difference of weights
[tex]\begin{gathered} W_{\text{air}}-W_{\text{water}}=\rho_wgV \\ V=\frac{W_{\text{air}}-W_{\text{water}}}{\rho_wg} \\ V=\frac{297N-263N}{1000\cdot9.8} \\ V=0.00347\text{ m}^3 \end{gathered}[/tex]Now for the object
[tex]\begin{gathered} \rho_{\text{obj}}=\frac{W_{\text{air}}}{gV} \\ \rho_{\text{obj}}=\frac{297}{9.8\cdot0.00347} \\ \rho_{\text{obj}}=8735.29kg/m^3 \end{gathered}[/tex]
