Answer
The volume of the steel can be $9.3\times 10^{-3}~m^3$
Work Step by Step
We can convert the volume of the gasoline $V_g$ to units of $m^3$ as:
$V_g = (210~L)(10^{-3}~m^3/L)$
$V_g = 0.21~m^3$
Let $V_s$ be the volume of the steel drum. The total volume of the drum is $V_s+V_g$. Let's assume that the drum is completely submerged in water and the buoyancy force is equal to the weight of the drum of gasoline. Let $M_s$ be the mass of the steel. Let $M_g$ be the mass of the gasoline. Let $\rho_w$ be the density of fresh water. Therefore;
$F_B = M_s~g+M_g~g$
$\rho_w~(V_s+V_g)~g = V_s~\rho_s~g+V_g~\rho_g~g$
$V_s~(\rho_s- \rho_w) =V_g~(\rho_w- \rho_g)$
$V_s = \frac{V_g~(\rho_w-\rho_g)}{\rho_s- \rho_w}$
$V_s = \frac{(0.21~m^3)[1.00\times 10^3~kg/m^3- 0.70\times 10^3~kg/m^3]}{7.8\times 10^3~kg/m^3- 1.00\times 10^3~kg/m^3}$
$V_s = 9.3\times 10^{-3}~m^3$
The volume of the steel can be $9.3\times 10^{-3}~m^3$.
