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The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is
Explanation
$\eta = 1 - T_2/T_1$ with $T_1 = 373$ K and $T_2 = 273$ K.
Detailed Solution
Efficiency of ideal heat engine, $\eta = \left(1 - \frac{T_2}{T_1}\right)$
$T_2$ : sink temperature, $T_1$ : source temperature
$\%\eta = \left(1 - \frac{T_2}{T_1}\right)\times 100 = \left(1 - \frac{273}{373}\right)\times 100$
$= \left(\frac{100}{373}\right)\times 100 = 26.8\%$