Work Done In Isothermal Expansion

As derived in NCERT Class 11, Chapter 12 (Thermodynamics), the work done by an ideal gas in a reversible isothermal expansion is $W=nRT\ln \left(\frac{V_f}{V_i}\right)$, where $n$ is the number of moles and the temperature is constant. Here $n=2$, $T=300$ K, and $\frac{V_f}{V_i}=\frac{2V}{V}=2$, so $\ln 2\approx 0.693$. Substituting: $W=2\times 8.3\times 300\times 0.693=4980\times 0.693\approx 3453$ J. Since temperature is constant, $$\begin{gathered} \Delta \\ cup=0 \end{gathered}$$ and all this work is supplied by heat absorbed from the surroundings. The option 4980 J is wrong because it omits the $\ln 2$ factor, using $nRT$ alone. The option 1726 J is wrong because it uses one mole instead of two. The option 6906 J is wrong because it doubles the correct value, perhaps using $\ln 4$ or a factor error. A plausibility check on magnitude confirms the answer: $nRT=2\times 8.3\times 300=4980$ J, and multiplying by $\ln 2<1$ must give a value below 4980 J, which the result of 3453 J indeed satisfies, so the magnitude and the logarithmic dependence are both reasonable.