The current flowing through $$R_2$$ is:

The left-hand branch contains a cell of emf $$E_1 = 6\ \text{V}$$ and internal resistance $$r_1 = 2\ \Omega$$.
The right-hand branch contains a cell of emf $$E_2 = 4\ \text{V}$$ and internal resistance $$r_2 = 1\ \Omega$$.
The two junctions of the branches are connected by the resistor $$R_2 = 3\ \Omega$$, whose current we have to find.

Case 1: Open-circuit condition ( $$R_2$$ removed )
With no external connection between the junctions, no current flows through either internal resistance, so the potentials of the junctions are simply equal to the emfs of their respective cells: the left junction is at $$6\ \text{V}$$ and the right junction at $$4\ \text{V}$$ with respect to the common return conductor.

Hence, when $$R_2$$ is connected, a potential difference of
$$V = E_1 - E_2 = 6\ \text{V} - 4\ \text{V} = 2\ \text{V}$$
exists across it, driving current from the left junction (higher potential) to the right junction (lower potential).

Case 2: Closed circuit ( $$R_2$$ in place )
The three resistances $$r_1,\, R_2$$ and $$r_2$$ now form one continuous path for the current. Applying Ohm’s law to this single loop,

$$I = \frac{\text{net emf}}{\text{total resistance}}$$

The net emf acting in the direction of current is $$E_1 - E_2 = 2\ \text{V}$$, and the total resistance encountered is
$$R_{\text{total}} = r_1 + R_2 + r_2 = 2\ \Omega + 3\ \Omega + 1\ \Omega = 6\ \Omega$$.

Therefore,
$$I = \frac{2\ \text{V}}{6\ \Omega} = \frac{1}{3}\ \text{A}$$.

The current of $$\tfrac{1}{3}\ \text{A}$$ flows from the left cell to the right cell through $$R_2$$, that is, from the 6 V side to the 4 V side.

Hence, the magnitude of the current through $$R_2$$ is $$\tfrac{1}{3}\ \text{A}$$, which matches Option C.