C

Step 1: Identify the temperatures involved. The high temperature reservoir (T_H) is 900K and the low temperature reservoir (T_C) is 300K.
Step 2: Use the efficiency formula of a Carnot engine, which is given by:
\[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} \]
Step 3: The work done (W) by the Carnot engine can be expressed in terms of heat absorbed (Q_H) from the hot reservoir:
\[ W = \eta Q_H \]
Step 4: Given that the heat rejected (Q_C) to the cold reservoir relates to work done as follows:
\[ Q_C = Q_H - W = Q_H - \eta Q_H = Q_H \left( 1 - \eta \right) = Q_H \left( \frac{1}{3} \right) \]
Step 5: Since we know the heat rejected (Q_C) in this context is 1200J, we can find Q_H:
\[ \frac{1}{3} Q_H = 1200 \Rightarrow Q_H = 1200 \times 3 = 3600J \]
Step 6: Now we can use the heat capacity relation to find heat delivered to the low-temperature reservoir:
\[ Q_C = \frac{T_C}{T_H} Q_H = \frac{300}{900} Q_H = \frac{1}{3} Q_H = 1200J \]
Therefore, the heat energy delivered by the engine to the low temperature reservoir, in one cycle, is 1200J.