A heat engine has an efficiency of $$\frac{1}{6}$$. When the temperature of sink is reduced by 62°C, its efficiency gets doubled. The temperature of the source is:

Let us denote the temperature of the hot reservoir (source) by $$T_1$$ and that of the cold reservoir (sink) by $$T_2$$. Throughout the calculation all temperatures will be expressed on the absolute (kelvin) scale; a change of 1 °C equals a change of 1 K, so differences can be handled in either unit without confusion.

For an ideal (Carnot) heat engine, the efficiency $$\eta$$ is given by the well-known relation

$$\eta \;=\;1-\dfrac{T_2}{T_1}.$$

We are first told that the efficiency is $$\dfrac16$$. Substituting this value into the formula we have

$$1-\dfrac{T_2}{T_1}\;=\;\dfrac16.$$

Rearranging term by term,

$$\dfrac{T_2}{T_1}\;=\;1-\dfrac16\;=\;\dfrac56,$$

and hence

$$T_2=\dfrac56\,T_1. \quad -(1)$$

Next, the problem states that the sink temperature is lowered by 62 °C (that is, by 62 K). Let the new sink temperature be $$T_2'$$. Then

$$T_2' = T_2 - 62.$$

With this altered sink temperature the efficiency doubles, becoming $$2\times\dfrac16=\dfrac13$$. Applying the efficiency formula once more, we get

$$1-\dfrac{T_2'}{T_1}\;=\;\dfrac13.$$

Solving for the ratio,

$$\dfrac{T_2'}{T_1}\;=\;1-\dfrac13\;=\;\dfrac23,$$

so

$$T_2'=\dfrac23\,T_1. \quad -(2)$$

Now we substitute $$T_2=\dfrac56\,T_1$$ from equation (1) into the definition of $$T_2'$$:

$$T_2' = T_2 - 62 \;=\;\dfrac56\,T_1\;-\;62.$$

But equation (2) tells us that $$T_2'=\dfrac23\,T_1$$. Equating the two expressions for $$T_2'$$ gives

$$\dfrac56\,T_1\;-\;62 \;=\;\dfrac23\,T_1.$$

To collect like terms, we bring the right-hand term to the left:

$$\dfrac56\,T_1 - \dfrac23\,T_1 \;=\;62.$$

Writing both fractions with a common denominator 6, $$\dfrac23\,T_1=\dfrac46\,T_1$$, so

$$\left(\dfrac56 - \dfrac46\right)T_1 \;=\;62.$$

The difference in the parentheses is $$\dfrac16$$, hence

$$\dfrac16\,T_1 \;=\;62.$$

Multiplying both sides by 6 we arrive at

$$T_1 \;=\;372\;\text{K}.$$

Finally, converting back to the Celsius scale (recall $$T(\text{°C}) = T(\text{K}) - 273$$), we obtain

$$T_1 = 372\;\text{K} - 273 = 99\;^\circ\text{C}.$$

Hence, the correct answer is Option D.