Four 2 cm × 2 cm × 2 cm cubes of ice are removed from a...

Volume of the four ice cubes is $V_{\text{ice}} = 4 \times (2 \times 2 \times 2) = 32 \text{ cm}^3 = 32 \times 10^{-6} \text{ m}^3$.

Mass of the ice is $m_{\text{ice}} = V_{\text{ice}} \times \rho_{\text{ice}} = 32 \times 10^{-6} \times 900 = 0.0288 \text{ kg}$.

Mass of the drink is $m_{\text{drink}} = V_{\text{drink}} \times \rho_{\text{drink}} = 200 \times 10^{-6} \times 1000 = 0.2 \text{ kg}$.

Assuming the initial temperature of the ice is 0^{}^{{}^{\circ}} \text{C}, the heat required to melt all the ice is $Q_{\text{req}} = m_{\text{ice}} \times L = 0.0288 \times 3.4 \times 10^5 = 9792 \text{ J}$.

The maximum heat that can be released by the drink in cooling from 10^{}^{{}^{\circ}} \text{C} to 0^{}^{{}^{\circ}} \text{C} is $Q_{\text{rel}} = m_{\text{drink}} \times c_{\text{drink}} \times \Delta T = 0.2 \times 4200 \times 10 = 8400 \text{ J}$.

Since $Q_{\text{rel}} < Q_{\text{req}}$, the heat released by the drink is insufficient to melt all the ice. Thus, only a fraction of the ice melts, and the mixture will consist of both ice and liquid.

Therefore, the system reaches thermal equilibrium at the melting point of ice, which is 0^{}^{{}^{\circ}} \text{C}.