In the circuit as shown in figure the

$$\begin{gathered} \frac{1}{R^{\mathrm{\prime }}}=\frac{1}{10}+\frac{1}{10}+\frac{1}{20}\Rightarrow R^{\mathrm{\prime }}=\frac{20}{5}=4\mathrm{\Omega } \\ \mathrm{Now} \mathrm{using} \mathrm{ohms} \mathrm{law}, i=\frac{25}{R+R^{\mathrm{\prime }}}\Rightarrow 0.5=\frac{25}{R+4} \end{gathered}$$

$\Rightarrow R+4=\frac{25}{0.5}=50\Rightarrow R=50-4=46\mathrm{\Omega }$

Current through 20Ω resistor

$=\frac{0.5\times 5}{20+5}=\frac{2.5}{25}=0.1\mathrm{A}$

Potential difference across middle resistor

= Potential difference across

$20\mathrm{\Omega }=20\times 0.1=2V$