Answer
$\rho = 1.8~g/cm^3$
Work Step by Step
After the depth reaches $h = 1.5~cm$, the apparent weight does not change. This must be because the entire block is submerged after $h=1.5~cm$. Thus $d = 1.5~cm$
We can find the buoyancy force when the block is completely submerged:
$W_{app} = W-F_b$
$F_b = W - W_{app}$
$F_b = (0.25~N)-(0.10~N)$
$F_b = 0.15~N$
The buoyancy force is equal to the weight of the displaced liquid. We can find the mass of the liquid:
$mg = F_b$
$m = \frac{F_b}{g}$
$m = \frac{0.15~N}{9.8~m/s^2}$
$m = 0.0153~kg$
We can find the density of the liquid:
$\rho = \frac{m}{V}$
$\rho = \frac{m}{d~A}$
$\rho = \frac{15.3~g}{(1.5~cm)(5.67~cm^2)}$
$\rho = 1.8~g/cm^3$
