For an ideal gas expanding isothermally from volume $V_1$ to $V_2$ at temperature $T$, the work done by the gas is:

For an isothermal process, $P = \dfrac{nRT}{V}$. Work done by the gas equals $\int_{V_1}^{V_2} P\,dV = nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = nRT \ln\dfrac{V_2}{V_1}$. If $V_2 > V_1$ (expansion) the log is positive and the gas does positive work, as it should. Option D flips the limits and gives the wrong sign.