The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the temperature of the hot reservoir and $T_2$ is the temperature of the cold reservoir (in Kelvin). Here, $T_1 = 1000$ K and $T_2 = 300$ K.
\[
\eta = 1 - \frac{300}{1000} = 1 - 0.3 = 0.7
\]
The efficiency in percentage is $0.7 \times 100 = 70%$.