Question

The interior of a refrigerator has a surface area of 4.8 m2. It is insulated by a 2.2 cm thick material that has a thermal conductivity of 0.0115 J/m · s · ◦ C. The ratio of the heat extracted from the interior to the work done

Answer 1

The question is incomplete, if complete question contains

"by the motor is 7.7% of the theoretical maximum. The temperature of the room is 33.3°C, and the temperature inside the refrigerator is 6.3°C.
Determine the power required to run the compressor. Answer in units of W", with it,

Then answer will be....

Some heat flows from room through insulation into the refrigerator.
You can calculate this amount using Fourier's law:
H = ∆Q/∆t = k∙A∙∆T/∆x
= 0.0115J/ms°C × 4.2m² ×(33.3°C - 6.8°C) / 0.022m
= 58.18J/s = 58.18W

The fridge system has to remove this amount from interior in order to keep internal temperature constant.

The coefficient of performance(COP) of a fridge is the ratio of heat removed to work done, or heat removed per unit time, to power input:
COP = Q/W = H/P

The maximum coefficient of performance of a fridge is
COP(max) = T(cool)/(T(hot) -T(cool)

Here
T(cool)= 6.8°C = 279.95K
T(Hot) = 33.3°C = 306.45K
=>
COP(max) = 279.95K/(306.45K - 279.95K) = 10.564

The actual COP of the fridge is 7,7% of that maximum value.i.e.
COP = 0.077 ×COP_max = 0.813

Hence:
P = H/COP
= 58.18W / 0.813
= 71.56W