A tub is filled with water and a wooden cube 10 cm × 10 cm × 10 cm...

Let $A$ be the base area of the wooden cube.

$A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2$

Let $\Delta x$ be the increase in the submerged depth of the cube when the coin is placed on it.

$\Delta x = 3.87 \text{ cm}$

The additional volume of water displaced due to the weight of the metal coin is $\Delta V = A \times \Delta x$.

$\Delta V = 100 \times 3.87 = 387 \text{ cm}^3$

By Archimedes' principle, the weight of the metal coin is balanced by the additional buoyant force, which is equal to the weight of the additional water displaced.

$m g = \Delta V \rho_w g$

$m = \Delta V \rho_w$

Substituting the values of $\Delta V$ and the density of water $\rho_w = 1 \text{ g/cm}^3$:

$m = 387 \text{ cm}^3 \times 1 \text{ g/cm}^3 = 387 \text{ g}$

Answer: $387$