Hypothetical Hydrostatics

For a compressible fluid whose density varies with depth, we cannot apply the pressure formula of $p = \rho(h)g h$ or ( $\rho(y)g y$ in this problem) directly. To find the pressure $p(y)$ at a depth of $y$ from the liquid surface, consider a column of the liquid of cross-sectional area $A$ on top of a point at depth $y$ . The weight of the column of liquid at this point is given by:

$F_g = \int_0^y \rho(z) A \ \text dz \cdot g = g A \int_0^y \rho_0 z \ \text dz = \frac {\rho_0 g A y^2}2$

The pressure at a depth of $y$ is:

$p(y) = \frac {\rho_0 g A y^2}{2A} = \frac {\rho_0 g y^2}2$

And the torque exerted on wall $CB$ about $C$ is:

$T = \int_0^h p(y) w \ \text dy \cdot y = \int_0^h \frac {\rho_0 g w y^3}2\ \text dy = \frac {\rho_0 g w h^4}8$

Therefore $ab = 4\cdot 8 = \boxed{32}$