Since the given resistance is in series, so, total resistance $\left(R\right)=20+30=50 \mathrm{\Omega }$

Given,

Potential difference$=25 \mathrm{V}$

Current$\left(I\right)=\frac{V}{R}=\frac{25}{50}=0.5 \mathrm{A}$

Also, Potential gradient $=\frac{{\mathrm{V}}_{\mathrm{pd}}}{L}=\frac{IR}{L}$

Where, $V_{pd}=$potential drop at wire

$L=$Length of wire

Potential gradient$=\frac{20\times 0.5}{10}=1 \mathrm{V} {\mathrm{m}}^{-1}$

So, Emf of the cell is

$E=$Potential gradient$\times$Balancing Length

$E= 1\times \frac{250}{100}$

$E= 2.5 \mathrm{V}=\frac{25}{10} \mathrm{V}$

Thus, $x=25$.