In the figure shown, the current in the $$10\,\text{V}$$ battery is close to:

Let the bottom-middle junction be the reference ground ($$0\text{ V}$$).

$$\text{Potential of the top-right node } = V_1$$

$$\text{Potential of the bottom-right node } = V_1 - 10$$

Applying KCL at node $$V_1$$:

$$\frac{V_1 - 20}{5 + 2} + \frac{V_1 - 0}{10} + \frac{(V_1 - 10) - 0}{4} = 0$$

$$\frac{V_1 - 20}{7} + \frac{V_1}{10} + \frac{V_1 - 10}{4} = 0$$

$$20(V_1 - 20) + 14V_1 + 35(V_1 - 10) = 0$$

$$20V_1 - 400 + 14V_1 + 35V_1 - 350 = 0 \implies 69V_1 = 750 \implies V_1 = \frac{750}{69} \approx 10.87\text{ V}$$

Current through the $$10\text{ V}$$ battery branch (from top to bottom):  $$I = \frac{(V_1 - 10) - 0}{4} = \frac{10.87 - 10}{4} = \frac{0.87}{4} \approx 0.217\text{ A}$$