A cubical block of ice floating in water must support a metal piece...

Let the edge length of the cubical block of ice be $L$.

Volume of the ice block, $V = L^3$

Density of water, $\rho_{\text{w}} = 1000 \text{ kg/m}^3$

Density of ice, $\rho_{\text{ice}} = 0.9 \times 1000 = 900 \text{ kg/m}^3$

Mass of the ice block, $m_{\text{ice}} = V \rho_{\text{ice}} = 900 L^3$

Mass of the metal piece, $m_{\text{metal}} = 0.5 \text{ kg}$

For the ice block to just float without sinking, it must be completely submerged. The buoyant force provided by the displaced water must balance the total weight of the ice block and the metal piece.

Equating the buoyant force to the total weight:

$V \rho_{\text{w}} g = (m_{\text{ice}} + m_{\text{metal}}) g$

$1000 L^3 = 900 L^3 + 0.5$

$100 L^3 = 0.5$

$L^3 = 0.005 \text{ m}^3$

Converting the volume to $\text{cm}^3$:

$L^3 = 0.005 \times 10^6 \text{ cm}^3 = 5000 \text{ cm}^3$

$L = (5000)^{1/3} \text{ cm} \approx 17.1 \text{ cm}$

The closest value among the given options is $17 \text{ cm}$.