#### Answer

The block's mass density is $667~kg/m^3$

#### Work Step by Step

We can find the volume $V_s$ of the part of the block that is submerged as:
$V_s = (0.020~m)(0.020~m)(0.040~m)$
$V_s = 1.6\times 10^{-5}~m^3$
The buoyant force on the block will be equal to the block's weight. The buoyant force is equal to the weight of the water which is displaced. Let $\rho_w$ be the density of the water. We can find the mass $M$ of the block as:
$M~g = F_B$
$M~g = \rho~V_s~g$
$M = \rho~V_s$
$M = (1000~kg/m^3)(1.6\times 10^{-5}~m^3)$
$M = 0.016~kg$
We can find the density $\rho_b$ of the block as:
$\rho_b = \frac{M}{V}$
$\rho_b = \frac{0.016~kg}{(0.020~m)(0.020~m)(0.060~m)}$
$\rho_b = 667~kg/m^3$
The block's mass density is $667~kg/m^3$.