A Carnot engine has an efficiency of 50% when its sink is at 300text{ K}. If the efficiency is to be increased to 70% keeping the source temperature…

Initial efficiency $\eta_1 = 1 - \frac{T_2}{T_1} \implies 0.5 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.5 \implies T_1 = 600\text{ K}$. New efficiency $\eta_2 = 0.7$, source temperature $T_1 = 600\text{ K}$, let new sink temperature be $T_2'$. Then $0.7 = 1 - \frac{T_2'}{600} \implies \frac{T_2'}{600} = 0.3 \implies T_2' = 180\text{ K}$. Decrease in sink temperature = $300 - 180 = 120\text{ K}$.