Given initially a resistance $P = 4 \mathrm{\Omega }$ is connected in the left gap and let $Q$ is connected in the right gap of meter bridge and the null point is obtained at $60 \mathrm{cm}$,

From the balanced condition of the Wheatstone bridge 

$\frac{P}{Q}=\frac{R}{S} ...\left(1\right)$

Let $\rho$ is the resistance per unit length of meter bridge wire than

For wire $AN$, $P=\rho l=40 \rho \left(\because l = 40 \mathrm{cm}\right)$ 

And for NB wire 

$$\begin{gathered} Q = \rho \left(100-l\right) \\ Q= \rho \left(100-60\right)=40 \rho \end{gathered}$$  

On substituting the values from equation $\left(1\right)$

$\frac{4}{60}=\frac{Q}{40}$

$\Rightarrow Q=\frac{16}{6}=\frac{8}{3} \mathrm{\Omega }$

Now an unknown resistance $R$ is connected in series with the $4 \mathrm{\Omega }$ resistance and the null point is obtained at $l=80 \mathrm{cm}$ 

Now, $P = \rho l =80 l$

and $$\begin{gathered} Q = \rho \left(100-l\right) \\ Q= \rho \left(100-80\right)= 20 \rho \end{gathered}$$

Then from equation $\left(1\right)$

 $$\begin{gathered} \frac{4+R}{80}=\frac{Q}{20} \\ \Rightarrow \frac{4+R}{80}=\frac{{\displaystyle \frac{8}{3}}}{20} \left(\because Q=\frac{8}{3}\right) \\ \Rightarrow \frac{4+R}{80}=\frac{8}{60} \\ \Rightarrow R = \frac{20}{3} \mathrm{\Omega } \end{gathered}$$