Answer
The book asks us to show that 3000 cal are required to raise the temperature of 300 g of water from $22^{\circ}C$ to $30^{\circ}C$ . There is a typo; the answer is actually 2400 calories.
Work Step by Step
$$Q = cm \Delta T$$
$$=\frac{1 cal}{g \cdot ^{\circ}C}(300 g)(30^{\circ}C - 22^{\circ}C ) = 2400 cal$$
