Given circuit is a parallel combination of two cells connected across a resistance $R$.

Now, for parallel combination of cells

$V_{ eq }=\frac{\frac{V_1+V_2}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}$

$\Rightarrow \quad V_{ eq }=\frac{V_1 R_2+V_2 R_1}{R_1+R_2} \quad \dots(i)$

$\text { and } R_{ eq }=\frac{R_1 R_2}{R_1+R_2} \quad \dots(ii)$

So, current through resistance $R$ is

$I=\frac{V_{\text {eq }}}{R+R_{\text {eq }}} \quad \dots(iii)$

In case $A$

$V_1=2 V , V_2=0 V \text { and } I=3 \,mA$

So, from Eqs. $(i), (ii)$ and $(iii)$, we get

$V_{\text {eq }}=\frac{2 \cdot R_2+0 \cdot R_1}{R_1+R_2}=\frac{2 R_2}{R_1+R_2}$

$R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2} \text { and } I=\frac{2 R_2 / R_1+R_2}{R+\frac{R_1 R_2}{R_1+R_2}}$

$\Rightarrow \quad 3 mA =\frac{2 R_2}{R\left(R_1+R_2\right)+R_1 R_2}$

In case $B$

$V_1=0 \,V , V_2=4 \,V \text { and } I=4 \,mA$

$\Rightarrow \quad 4 \,mA =\frac{4 R_1}{R\left(R_1+\overline{R_2}\right)+R_1 R_2} \quad \dots(v)$

From Eqs. $(iv)$ and $(v)$, we have

$\frac{3}{4}=\frac{2 R_2}{4 R_1} \Rightarrow \frac{3}{4}=\frac{R_2}{2 R_1} \Rightarrow \frac{R_2}{R_1}=\frac{3}{2}$

In case $C$

$V_1=10 \,V$ and $V_2=10 \,V$

So, if current through resistance $R$ is $I$, then

$V_{e q}=\frac{10 R_1+10 R_2}{R_1+R_2}$

and so $I=\frac{10 R_1+10 R_2}{R\left(R_1+R_2\right)+R_1 R_2} \quad \dots(vi)$

Now, from Eqs. $(iv)$ and $(vi)$, we have

$\frac{3}{I}=\frac{\left(\frac{2 R_2}{R\left(R_1+R_2\right)+R_1 R_2}\right)}{\left(\frac{10 R_1+10 R_2}{R\left(R_1+R_2\right)+R_1 R_2}\right)}$

$\Rightarrow \quad \frac{3}{I}=\frac{2 R_2}{10 R_1+10 R_2}$

$\Rightarrow \quad \frac{I}{3}=5 \frac{R_1}{R_2}+5$

$\Rightarrow \quad \frac{I}{3}=5\left(\frac{2}{3}\right)+5$

$\Rightarrow \quad I=10+15=25 \,mA$