ANSWER
Option A:
[tex]A.\text{ }(1-\frac{H_{above}}{H_{total}})(\rho_{flu\imaginaryI d})[/tex]
EXPLANATION
We have to solve the formula for ρ_object. To do so, first, divide both sides by H_total,
[tex]\begin{gathered} \frac{H_{total}*(1-\frac{\rho_{objet}}{\rho_{fluid}})}{H_{total}}=\frac{H_{above}}{H_{total}} \\ \\ \left(1-\frac{\rho_{object}}{\rho_{fluid}}\right)=\frac{H_{above}}{H_{total}} \end{gathered}[/tex]
Then, subtract 1 from both sides,
[tex]\begin{gathered} 1-1-\frac{\rho_{object}}{\rho_{fluid}}=\frac{H_{above}}{H_{total}}-1 \\ \\ \begin{equation*} -\frac{\rho_{object}}{\rho_{fluid}}=\frac{H_{above}}{H_{total}}-1 \end{equation*} \end{gathered}[/tex]
And then, multiply both sides by -ρ_fluid,
[tex]\begin{gathered} -\frac{\rho_{object}}{\rho_{fluid}}\cdot(-\rho_{fluid})=\left(\frac{H_{above}}{H_{total}}-1\right)\cdot(-\rho_{fluid}) \\ \\ \rho_{object}=\left(1-\frac{H_{above}}{H_{total}}\right)\cdot(\rho_{fluid}) \end{gathered}[/tex]
Hence, the formula for ρ_object is,
[tex]\rho_{object}=(1-\frac{H_{above}}{H_{total}})(\rho_{flu\imaginaryI d})[/tex]