Answer
$6.6\times10^3kcal$
Work Step by Step
Eighty percent of the heat from the cyclist causes the water to evaporate. The water is at approximately room temperature, so use the heat of vaporization at room temperature as discussed on page 399. One liter of water has a mass of 1 kg.
$$0.80Q_{cyclist}=mL_v$$
$$ Q_{cyclist}=\frac{mL_v }{0.8}=\frac{(9.0kg)(585kcal/kg)}{0.80}=6.6\times10^3kcal$$