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(C) Let $R_{eq} = R_g + R$ be the total external resistance. Let $E = 1.5\, V$ be the $EMF$ of each battery and $r$ be the internal resistance.Case $I

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$r = \frac{1}{3}\, \Omega$

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(C) Let $R_{eq} = R_g + R$ be the total external resistance. Let $E = 1.5\, V$ be the $EMF$ of each battery and $r$ be the internal resistance.Case $I$: Batteries in series.The total $EMF$ is $2E = 3\, V$ and total internal resistance is $2r$.The current is $I_1 = \frac{2E}{R_{eq} + 2r} = 1\, A$.$R_{eq} + 2r = 3$  --- $(1)$Case $II$: Batteries in parallel.The total $EMF$ is $E = 1.5\, V$ and total internal resistance is $\frac{r}{2}$.The current is $I_2 = \frac{E}{R_{eq} + \frac{r}{2}} = 0.6\, A$.$R_{eq} + 0.5r = \frac{1.5}{0.6} = 2.5$  --- $(2)$Subtracting $(2)$ from $(1)$:$(R_{eq} + 2r) - (R_{eq} + 0.5r) = 3 - 2.5$$1.5r = 0.5$$r = \frac{0.5}{1.5} = \frac{1}{3}\, \Omega$.

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