Answer
$2.8\;\rm kW$
Work Step by Step
We know that electric power is given by
$$P=\dfrac{W_{in}}{\Delta t}\tag 1$$
And we know that the coefficient of performance for a Carnot refrigerator is given by
$$K_{\rm carnot}=\dfrac{T_C}{T_H-T_C}=\dfrac{Q_C}{W_{in}}$$
Hence,
$$W_{in}=\dfrac{Q_C}{K_{\rm carnot}}$$
$$W_{in}=\dfrac{Q_C(T_H-T_C)}{T_C}\tag 2$$
where $Q_C$ is the heat transferred from the cold reservoir to the system which is given by
$$\dfrac{Q_C}{\Delta t}=\dfrac{kA\Delta T}{x}$$
where $Q_C/\Delta t$ is the rate of heat transfer, $k$ is the thermal conductivity of stainless steel, $A$ is the area of the steel plate, $ x$ is the thickness of the steel plate, $T_2$ is the temperature inside the box ($-20^{\circ} \rm{C}$), and $T_1$ is the temperature of the room ($25^{\circ} \rm{C}$).
Thus,
$$ Q_C =\dfrac{kA(T_R-T_C)}{x}\Delta t$$
where $T_R$ is the room temperature.
Plugging into (2);
$$W_{in}=\dfrac{kA(T_R-T_C)}{x} \dfrac{ (T_H-T_C)}{T_C} \Delta t$$
Plugging into (1);
$$W_{in}=\dfrac{kA(T_R-T_C)}{x} \dfrac{ (T_H-T_C)}{T_C} \Delta t$$
$$P=\dfrac{kA(T_R-T_C)}{x} \dfrac{ (T_H-T_C)}{T_C} \color{red}{\bf\not}\Delta t\dfrac{ 1}{ \color{red}{\bf\not}\Delta t}$$
Plugging the known;
$$P=\dfrac{(14)(0.4\times 0.4)(25-[-20])}{0.01}\cdot \dfrac{ (50-[-20])}{(-20+273)} $$
$$P=\color{red}{\bf 2.8\times 10^3}\;\rm W$$
