In the case of potentiometer null point is obtained when the current flowing from the battery $E_1$ should be equal to current drawn from the battery $E_2$,

$$\begin{gathered} \Rightarrow i_1=i_2 \\ \Rightarrow \frac{E_1}{R_{wire}+R}=\frac{E_2}{R_{Balance}} \end{gathered}$$

here,

$E_1=5 \mathrm{V}$, $R_{wire}= 50 \Omega , R=450 \Omega \& E_2=E$

$\Rightarrow \frac{5}{50+450}=\frac{E}{R_{Balance}}$

the value of $R_{Balance}$is,

$R_{Balacne}=R_{wire}\times \frac{450}{1000}=50\times \frac{450}{1000}$

$$\begin{gathered} \Rightarrow \frac{5}{50+450}=\frac{E}{50\times \frac{450}{1000}} \\ \Rightarrow E=0.225 \mathrm{V} \end{gathered}$$