To solve this problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature.$\\$The formula for Newton's Law of Cooling is:$\\ \frac{dT}{dt} = -k(T - T_{room}) \\$Where:$\\$• $T$ is the temperature of the object at time $t. \\$• $T_{room}$ is the ambient (room) temperature.$\\$• $k$ is a positive constant.$\\$Given:$\\$• Initial temperature $T_1 = 90^{\circ}C,$ final temperature $T_2 = 80^{\circ}C. \\$• Room temperature $T_{room} = 20^{\circ}C. \\$• Time taken to cool from $90^{\circ}C$ to $80^{\circ}C$ is $t$ minutes.$\\$Using the formula:$\\ \frac{T_2 - T_{room}}{T_1 - T_{room}} = e^{-kt} \\$Substitute the known values:$\\ \frac{80 - 20}{90 - 20} = e^{-kt} \\ \frac{60}{70} = e^{-kt} \\ \frac{6}{7} = e^{-kt} \\$Now, for the second scenario:$\\$• Initial temperature $T_3 = 80^{\circ}C,$ final temperature $T_4 = 60^{\circ}C. \\$We need to find the time taken, say $t_2,$ for this cooling process.$\\$Using the same formula:$\\ \frac{T_4 - T_{room}}{T_3 - T_{room}} = e^{-kt_2} \\$Substitute the known values:$\\ \frac{60 - 20}{80 - 20} = e^{-kt_2} \\ \frac{40}{60} = e^{-kt_2} \\ \frac{2}{3} = e^{-kt_2} \\$Now, we have two equations:$\\ e^{-kt} = \frac{6}{7} \\ e^{-kt_2} = \frac{2}{3} \\$Taking the natural logarithm of both sides:$\\ -kt = \ln\left(\frac{6}{7}\right) \\ -kt_2 = \ln\left(\frac{2}{3}\right) \\$Divide the second equation by the first:$\\ \frac{-kt_2}{-kt} = \frac{\ln\left(\frac{2}{3}\right)}{\ln\left(\frac{6}{7}\right)} \\ t_2 = t \cdot \frac{\ln\left(\frac{2}{3}\right)}{\ln\left(\frac{6}{7}\right)} \\$Using logarithm properties, we approximate:$\\ \ln\left(\frac{2}{3}\right) \approx -0.4055 \\ \ln\left(\frac{6}{7}\right) \approx -0.1542 \\ t_2 = t \cdot \frac{-0.4055}{-0.1542} \\ t_2 \approx t \cdot \frac{13}{5} \\$Therefore, the time taken by a similar cup of coffee to cool from $80^{\circ}C$ to $60^{\circ}C$ is $\frac{13}{5} t. \\$This corresponds to Option 2.