The two cells oppose each other in the loop. Net emf: $$E_{net} = E_1 - E_2 = 6 - 4 = 2\ \text{V}$$ Total resistance: $$R = 1 + 3 = 4\ \Omega$$ Current: $$I = \frac{E_{net}}{R} = \frac{2}{4} = 0.5\ \text{A}$$ For the branch from B to C containing $E_2=4$ V and $3\,\Omega$: the potential difference between B and C is $$V_{BC} = E_2 + I r_2 = 4 + (0.5)(3) = 5.5\ \text{V}$$ (The current drives against $E_2$, so the drop across the $3\,\Omega$ adds to the cell emf.) Hence $V_{BC} = 5.5$ V.