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**Hint:**The Carnot efficiency can be calculated by using the equation: $\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}}$. As the temperatures of food inside the refrigerator and the temperature of the environment are given, mere substitution gives the solution.

**Formula used:**The Carnot efficiency,

\[\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}}\]

where ${T_C}$ is the temperature of the cold reservoir and ${T_H}$ is the temperature of the hot reservoir, both represented in absolute Kelvin.

The temperature given in Celsius scale can be converted into the Kelvin scale as follows:

\[{T_{Kelvin}} = {T_{Celcius}} + 273\]

**Complete step-by-step answer:**

It is given that the refrigerator works based on the Carnot cycle and thus we can consider it as a Carnot engine. We know that, according to Carnot's theorem, efficiency of a heat engine is dependent only on the temperatures of hot and cold reservoirs.

It is also given the temperature of the food material inside the refrigerator, that is the temperature of the cold reservoir, ${T_C} = 4^\circ C = 273 + 4 = 277K$.

Similarly, the temperature of the surrounding environment is the temperature of the hot reservoir, which can be written as:

\[{T_H} = 15^\circ C = 15 + 273 = 288K\]

Now, the Carnot efficiency of the refrigerator can be estimated as:

\[\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}} = 1 - \dfrac{{277}}{{288}}\]

Hence, on simplification we get the efficiency as:

\[\therefore \eta = 1 - 0.9618 = 0.0381 \approx 0.038\]

**So, the Carnot efficiency of the refrigerator is $0.038$ and the correct answer is option A.**

**Note:**All the heat engines working between two heat reservoirs can have an efficiency which is lesser than the efficiency of a Carnot engine. A Carnot engine works based on the second law of thermodynamics, according to which, the efficiency of a Carnot engine is independent of the nature of the materials.

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