A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

For a refrigerator working on the Carnot cycle, the relevant relation is the coefficient of performance (COP), written specifically for refrigeration as

$$\beta \;=\;\frac{\text{Heat extracted from the low-temperature reservoir }(Q_C)}{\text{Work input per cycle }(W)}.$$

For an ideal Carnot refrigerator operating between an absolute cold-reservoir temperature $$T_C$$ and a hot-reservoir temperature $$T_H,$$ the COP is also given by the thermodynamic formula

$$\beta \;=\;\frac{T_C}{T_H - T_C}.$$

We are told that the cold-reservoir temperature is $$T_C = 250\ \text{K}$$ and the hot-reservoir temperature is $$T_H = 300\ \text{K}.$$ Substituting these values into the COP formula, we obtain

$$\beta \;=\;\frac{250}{300 - 250} \;=\;\frac{250}{50} \;=\;5.$$

This numerical value of $$\beta$$ can now be connected to the first definition of COP. We have

$$\beta \;=\;\frac{Q_C}{W}.$$

The problem states that the refrigerator removes an amount of heat $$Q_C = 500\ \text{cal}$$ from the low-temperature reservoir during every cycle. We therefore write

$$5 \;=\;\frac{500\ \text{cal}}{W}.$$

To solve for the required work $$W,$$ we rearrange the equation algebraically:

$$W \;=\;\frac{500\ \text{cal}}{5}\;.$$

Since the answer choices are expressed in joules, we convert calories to joules using the conversion factor $$1\ \text{cal} = 4.2\ \text{J}.$$ First convert the 500 calories:

$$Q_C = 500\ \text{cal} \times 4.2\ \frac{\text{J}}{\text{cal}} = 2100\ \text{J}.$$

Now substitute $$Q_C = 2100\ \text{J}$$ into the work expression:

$$W \;=\;\frac{2100\ \text{J}}{5} = 420\ \text{J}.$$

Thus, the refrigerator requires 420 joules of work per cycle.

Hence, the correct answer is Option A.