The efficiency ($\eta$) of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.

Case 1: $\eta_1 = \frac{1}{6}$ $\frac{1}{6} = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$

Case 2: The sink temperature is reduced by $62^{\circ}\text{C}$ (which is a change of $62\text{ K}$), so the new sink temperature is $T_2' = T_2 - 62$. The efficiency doubles, so $\eta_2 = 2 \times \frac{1}{6} = \frac{1}{3}$. $\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$ $\frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3}$ $T_2 - 62 = \frac{2}{3}T_1$

Substitute $T_2 = \frac{5}{6}T_1$ into the second equation: $\frac{5}{6}T_1 - 62 = \frac{2}{3}T_1$ $\frac{5}{6}T_1 - \frac{4}{6}T_1 = 62$ $\frac{1}{6}T_1 = 62$ $T_1 = 372\text{ K}$

Convert to Celsius: $T_1 (^{\circ}\text{C}) = 372 - 273 = 99^{\circ}\text{C}$.