Answer
The true mass of the piece of aluminum is 4.0019 kg
Work Step by Step
Let $M$ be the true mass of the aluminum. Let $\rho_a$ be the density of aluminum. We can find an expression for the volume $V$ of the aluminum.
$V = \frac{M}{\rho_a}$
Let $M_a$ be the apparent mass of the aluminum. Let $\rho_A$ be the density of air. Note that the apparent weight is equal to the true weight of the aluminum minus the buoyancy force provided by the air.
$M~g-F_B = M_a~g$
$M~g-\rho_A~V~g = M_a~g$
$M-\rho_A~(\frac{M}{\rho_a}) = M_a$
$M~(1-\frac{\rho_A}{\rho_a}) = M_a$
$M = \frac{M_a~\rho_a}{\rho_a-\rho_A}$
$M = \frac{(4.0000~kg)(2.70\times 10^3~kg/m^3)}{2.70\times 10^3~kg/m^3-1.29~kg/m^3}$
$M = 4.0019~kg$
The true mass of the piece of aluminum is 4.0019 kg.
