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 temperature and the ambient temperature (surroundings).$\\$The formula for Newton's Law of Cooling is:$\\ \frac{dT}{dt} = -k(T - T_s) \\$where:$\\$• $T$ is the temperature of the object at time $t, \\$• $T_s$ is the temperature of the surroundings,$\\$• $k$ is a positive constant,$\\$• $\frac{dT}{dt}$ is the rate of change of temperature.$\\$Given:$\\$• Initial temperature $T_1 = 70^{\circ}C$ at $t = 0, \\$• Temperature after 5 minutes $T_2 = 60^{\circ}C, \\$• Temperature after another 5 minutes $T_3 = 54^{\circ}C. \\$Let's apply the formula for the first 5 minutes:$\\ \frac{T_2 - T_1}{5} = -k(T_{avg1} - T_s) \\$where $T_{avg1} = \frac{T_1 + T_2}{2} = \frac{70 + 60}{2} = 65^{\circ}C. \\$Rearranging gives:$\\ 60 - 70 = -5k(65 - T_s) \\ -10 = -5k(65 - T_s) \\ 2 = k(65 - T_s) \quad$(Equation 1)$\\$Now, apply the formula for the next 5 minutes:$\\ \frac{T_3 - T_2}{5} = -k(T_{avg2} - T_s) \\$where $T_{avg2} = \frac{T_2 + T_3}{2} = \frac{60 + 54}{2} = 57^{\circ}C. \\$Rearranging gives:$\\ 54 - 60 = -5k(57 - T_s) \\ -6 = -5k(57 - T_s) \\ \frac{6}{5} = k(57 - T_s) \quad$(Equation 2)$\\$From Equations 1 and 2, we have:$\\ 2 = k(65 - T_s) \\ \frac{6}{5} = k(57 - T_s) \\$Divide Equation 1 by Equation 2:$\\ \frac{2}{\frac{6}{5}} = \frac{k(65 - T_s)}{k(57 - T_s)} \\ \frac{10}{6} = \frac{65 - T_s}{57 - T_s} \\ \frac{5}{3} = \frac{65 - T_s}{57 - T_s} \\$Cross-multiply to solve for $T_s: \\ 5(57 - T_s) = 3(65 - T_s) \\ 285 - 5T_s = 195 - 3T_s \\ 285 - 195 = 5T_s - 3T_s \\ 90 = 2T_s \\ T_s = \frac{90}{2} \\ T_s = 45^{\circ}C \\$Therefore, the temperature of the surroundings is $45^{\circ}C. \\$This corresponds to Option 1.