#### Answer

The tension in the string is 1.87 N

#### Work Step by Step

We can find the mass of the aluminum.
$M = V~\rho_{Al}$
$M = (100\times 10^{-6}~m^3)(2700~kg/m^3)$
$M = 0.27~kg$
We can find the buoyant force on the aluminum. The buoyant force is equal to the weight of the Ethyl alcohol that is displaced by the volume of the aluminum. Let $\rho_E$ be the density of Ethyl alcohol.
$F_B = \rho_E~V~g$
$F_B = (790~kg/m^3)~(100\times 10^{-6}~m^3)~(9.80~m/s^2)$
$F_B = 0.7742~N$
The sum of the tension and the buoyant force is equal to the aluminum's weight. We can find the tension in the string.
$T+F_B = Mg$
$T = Mg-F_B$
$T = (0.27~kg)(9.80~m/s^2)- (0.7742~N)$
$T = 1.87~N$
Therefore, the tension in the string is 1.87 N