To solve this problem, we need to understand the working of a Carnot engine and its application as a refrigerator.$\\$Given:$\\$• Efficiency of the Carnot engine as a heat engine is $\frac{1}{10}. \\$• Work done on the system $W = 10$ J$. \\$The efficiency $\eta$ of a heat engine is given by:$\\ \eta = \frac{W}{Q_H} \\$where $Q_H$ is the heat absorbed from the hot reservoir.$\\$Given efficiency $\eta = \frac{1}{10},$ we have:$\\ \frac{1}{10} = \frac{W}{Q_H} \\$Substitute $W = 10$ J:$\\ \frac{1}{10} = \frac{10}{Q_H} \\$Solve for $Q_H: \\ Q_H = 10 \times 10 = 100$ J$\\$Now, when the Carnot engine is used as a refrigerator, the coefficient of performance (COP) is given by:$\\$COP$= \frac{Q_L}{W} \\$For a Carnot refrigerator, the COP is related to the efficiency of the Carnot engine by:$\\$COP$= \frac{1}{\eta} - 1 \\$Substitute $\eta = \frac{1}{10}: \\$COP$= \frac{1}{\frac{1}{10}} - 1 = 10 - 1 = 9 \\$Now, using the COP to find $Q_L: \\ Q_L =$COP$\times W \\ Q_L = 9 \times 10 = 90$ J$\\$Therefore, the amount of energy absorbed from the reservoir at lower temperature is $90$ J.$\\$This corresponds to Option 1.