To solve this problem, we need to understand the relationship between the efficiency of a Carnot engine and its operation as a refrigerator.$\\$Given:$\\$• Efficiency of the Carnot engine as a heat engine $\eta = \frac{1}{10}. \\$• Work done on the system $W = 10$ J$. \\$The efficiency $\eta$ of a Carnot engine is given by:$\\ \eta = \frac{W}{Q_H} \\$where $Q_H$ is the heat absorbed from the hot reservoir.$\\$Rearranging the formula for $Q_H: \\ Q_H = \frac{W}{\eta} \\$Substitute the given values:$\\ Q_H = \frac{10}{\frac{1}{10}} = 100$ J$\\$As a refrigerator, the Carnot cycle absorbs heat $Q_L$ from the cold reservoir and expels $Q_H$ to the hot reservoir.$\\$The relationship between $Q_H, Q_L,$ and $W$ is:$\\ Q_H = Q_L + W \\$Rearranging for $Q_L: \\ Q_L = Q_H - W \\$Substitute the known values:$\\ Q_L = 100 - 10 = 90$ J$\\$Therefore, the amount of energy absorbed from the reservoir at lower temperature is $90$ J.$\\$This corresponds to Option 2.