To solve this problem, we need to determine the power required by the refrigerator.$\\$Given:$\\$• The refrigerator works between $4^{\circ}C$ and $30^{\circ}C. \\$• Heat removed per second $Q = 600$ calories.$\\$• 1 calorie $= 4.2$ Joules.$\\$First, convert the temperatures to Kelvin:$\\ T_{cold} = 4 + 273 = 277$ K$\\ T_{hot} = 30 + 273 = 303$ K$\\$The coefficient of performance (COP) for a refrigerator is given by:$\\ COP = \frac{T_{cold}}{T_{hot} - T_{cold}} \\$Substitute the known values:$\\ COP = \frac{277}{303 - 277} \\ COP = \frac{277}{26} \\$Calculate the COP:$\\ COP \approx 10.65 \\$The power required (input work) is related to the heat removed and the COP by:$\\ W = \frac{Q}{COP} \\$Convert the heat removed to Joules:$\\ Q = 600 \times 4.2 = 2520$ J/s$\\$Substitute the values into the power equation:$\\ W = \frac{2520}{10.65} \\$Calculate the power:$\\ W \approx 236.5$ W$\\$Therefore, the power required is approximately $236.5$ W.$\\$This corresponds to Option 1.