Interpret the circuit

Since the figure is not visible here, this standard JEE-type question corresponds to two resistors and three cells arranged so that we need the potential difference between points $a$ and $b$.

Given: $R_1=1\,\Omega, \quad R_2=2\,\Omega$ $E_1=2\,\text{V}, \quad E_2=E_3=4\,\text{V}$

Use loop/current balance

For the usual arrangement of this problem, the current through the branch containing $R_1$ is determined by the net emf across it due to the opposing cells, and similarly for $R_2$.

The effective current in the circuit comes out to be $I=\frac{E_2-E_1}{R_1+R_2}=\frac{4-2}{1+2}=\frac{2}{3}\,\text{A}$

Find the potential difference between $a$ and $b$

The potential difference $V_{ab}$ is obtained by moving from $a$ to $b$ across one branch. Using the branch with source $E_3$ and resistor drop: $V_{ab}=E_3-I R_2$

Substituting values: $V_{ab}=4-\left(\frac{2}{3}\right)(2)=4-\frac{4}{3}=\frac{8}{3}\,\text{V}$

$V_{ab}\approx 2.67\,\text{V}$

Match with the options

$V_{ab}\approx 2.7\,\text{V}$

So the correct option is: B: $2.7$ V

Compare with stored answer

Stored correct answer is A: 3.3 V.

My derived answer is B: 2.7 V. Hence I do not agree with the stored answer.

It appears the stored answer may be incorrect, unless the unseen circuit orientation is different from the standard interpretation. Under the usual interpretation consistent with these values, the result is $2.7\,\text{V}$.