The efficiency ($\eta$) of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$ where $T_1$ is the temperature of the source (hot reservoir) and $T_2$ is the temperature of the sink (cold reservoir) in Kelvin.

1. Identify Temperatures: Freezing point of water ($T_2$) = $0^{\circ}\text{C} = 273.15 \text{ K} \approx 273 \text{ K}$. Boiling point of water ($T_1$) = $100^{\circ}\text{C} = 373.15 \text{ K} \approx 373 \text{ K}$.

2. Calculate Efficiency: $\eta = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$ $\eta \approx 0.268$

3. Convert to Percentage: Percentage Efficiency = $0.268 \times 100\% = 26.8\%$.