The combination of two identical cells, whether connected in series...

When the cells are connected in series ${i}_{s} = \dfrac{2ε}{2r + R} = \dfrac{2ε}{2r + 2}$

When they are connected in parallel, the emf is ${ε}_{p} = \dfrac{\dfrac{ε}{r} + \dfrac{ε}{r}}{\dfrac{1}{r} + \dfrac{1}{r}} = ε$ and the effective internal resistance will be ${r}_{p} = \dfrac{r \times r}{r + r} = \dfrac{r}{2}$. Therefore, the current will be,

${i}_{p} = \dfrac{ε}{R + \dfrac{r}{2}} = \dfrac{ε}{2 + \dfrac{r}{2}} = \dfrac{2ε}{4 + r}$

Since, ${i}_{s} = {i}_{p} ⇒ \dfrac{2ε}{2r + 2} = \dfrac{2ε}{4 + r} ⇒ r = 2Ω$