Carnot’s cycle is said to have 25% efficiency when it operates between, T ( Source) and 300 K ( Sink) temperature T is  

$\begin{array}{l}\eta =\frac{T_2-T_1}{T_2},T_1=\text{ temperature of sink and }\\ T_2=\text{ temperature of source }\\ 0.25=\frac{T_2-T_1}{T_2}=\frac{T_2-300}{T_2}\\ 0.25T_2=T_2-300\\ 300=T_2-0.25T_2=0.75T_2\\ T_2=\frac{300}{0.75}=400K\end{array}$