#### Answer

The cooler must remove heat at 203 Joules per second.

#### Work Step by Step

Using the conductivity of Styrofoam (.029) as well as the surface area of the cooler (which is 35 square meters), we can find the rate at which heat enters the cooler. This is the same as the rate at which the cooler must remove heat:
$ H = -\frac{kA\Delta T}{\Delta x} = \frac{.029 \times 35 \times 16}{.08}=-203 \ W$