The work done on the system when one mole of an ideal gas at $500\mathrm{ }\mathrm{K}$ is compressed isothermally and reversibly to $1/10$ th of its original volume $(\mathrm{R}=2\mathrm{cal})$

For isothermal process,

$$\begin{gathered} \mathrm{W}=2.303 \mathrm{nRTlog}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1} \\ \mathrm{W}=2.303 (1 )\left(2\right) \left(500\right) \log \frac{10\mathrm{V}}{\mathrm{V}} \\ \mathrm{W}=2.303 \left(1\right) \left(2\right) (500 )\log 10 \\ \mathrm{W}=2303 \mathrm{cal} \\ \mathrm{W}=2.303 \mathrm{kcal} \end{gathered}$$

Thus, the required work done is $2.303\mathrm{kcal}$.

Hence, option (D) is correct.