Answer is: 13500 cal.
m(water) = V(water)·d(water) = 300ml·1g/ml = 300g.
ΔT = 45°C, temperature change.
Calorie is amount of energy (heat) needed to increase the temperature of one gram of water by 1°C.
To raise the temperature of 300 grams of water for 45°C:
Heat in calories = mass in gram x temperature change °C x specific heat (cal / gram·°C). Q = m·ΔT·c.
Q = 300g · 45°C · 1cal/g·°C = 13500 cal.