b
Potential difference across \(AB\) will be equal to battery equivalent across \(CD\)

\( V_{A B} =V_{C D}=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}+\frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}}=\frac{\frac{1}{1}+\frac{2}{1}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}} \)

\(=\frac{6}{3}=2\,V\)