#### Answer

(a) The volume of water is $6.32~m^3$
(b) The pressure at the bottom of the tank is $1.19\times 10^5~Pa$

#### Work Step by Step

(a) We can find the mass of the ethyl alcohol as:
$M = \rho_a~V$
$M = (790~kg/m^3)(8.0~m^3)$
$M = 6320~kg$
We can find the volume of the water with this mass as:
$V = \frac{M}{\rho_w}$
$V = \frac{6320~kg}{1000~kg/m^3}$
$V = 6.32~m^3$
The volume of water is $6.32~m^3$.
(b) If the water tank is a cube, we can find the length $L$ of each side.
$L = V^{1/3}$
$L = (6.32~m^3)^{1/3}$
$L = 1.85~m$
We can find the pressure at the bottom of the tank.
$P = P_0 + \rho~g~h$
$P = 1.01\times 10^5~Pa + (1000~kg/m^3)(9.80~m/s^2)(1.85~m)$
$P = 1.19\times 10^5~Pa$
The pressure at the bottom of the tank is $1.19\times 10^5~Pa$.