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(D) Let $E = 1.5\, V$ be the $EMF$ of each battery and $r$ be the internal resistance of each battery.When batteries are in series,total $EMF$ = $2E$

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$1/3\, \Omega$

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(D) Let $E = 1.5\, V$ be the $EMF$ of each battery and $r$ be the internal resistance of each battery.When batteries are in series,total $EMF$ = $2E$ and total internal resistance = $2r$.The current $I_1 = \frac{2E}{2r + R} = 1\, A$.So,$2E = 2r + R \Rightarrow 3 = 2r + R \Rightarrow R = 3 - 2r \quad (1)$When batteries are in parallel,total $EMF$ = $E$ and total internal resistance = $r/2$.The current $I_2 = \frac{E}{r/2 + R} = 0.6\, A$.So,$E = 0.6(r/2 + R) \Rightarrow 1.5 = 0.3r + 0.6R$.Multiplying by $10/3$,we get $5 = r + 2R \quad (2)$Substituting $(1)$ into $(2)$: $5 = r + 2(3 - 2r) = r + 6 - 4r = 6 - 3r$.$3r = 1 \Rightarrow r = 1/3\, \Omega$.

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