Efficiency of reversible engine is $\eta =\frac{1}{4}=\frac{W}{Q_{\mathrm{in}}}=1-\frac{T_2}{ T_1}$

From above, $\frac{T_2}{ T_1}=\frac{3}{4}$

When the temperature of the sink is reduced by $52 \mathrm{K}$ then its efficiency is doubled.

Then, $\frac{T_2-52}{ T_1}=\frac{1}{2}$

$$\begin{gathered} \Rightarrow \frac{T_2}{ T_1}-\frac{52}{ T_1}=\frac{1}{2} \\ \Rightarrow \frac{3}{4}-\frac{52}{ T_1}=\frac{1}{2} \\ \Rightarrow \frac{52}{ T_1}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4} \\ \Rightarrow T_1=52\times 4=208 \mathrm{K} \end{gathered}$$