A reversible heat engine converts one-fourth of the heat input into...

Efficiency of reversible engine is $η = \dfrac{1}{4} = \dfrac{W}{{Q}_{in}} = 1 - \dfrac{{T}_{2}}{{T}_{1}}$

From above, $\dfrac{{T}_{2}}{{T}_{1}} = \dfrac{3}{4}$

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

Then, $\dfrac{{T}_{2} - 52}{{T}_{1}} = \dfrac{1}{2}$

$⇒ \dfrac{{T}_{2}}{{T}_{1}} - \dfrac{52}{{T}_{1}} = \dfrac{1}{2} ⇒ \dfrac{3}{4} - \dfrac{52}{{T}_{1}} = \dfrac{1}{2} ⇒ \dfrac{52}{{T}_{1}} = \dfrac{3}{4} - \dfrac{1}{2} = \dfrac{1}{4} ⇒ {T}_{1} = 52 \times 4 = 208K$