Let current through resistor $X$ is be $i$.

Now, we consider section $d d$

Equating potential across $d d$, we get

$i_2 R=i_1(2 R)$

$\text { or } i_2=2 i_1$

Hence, current $i$ is

$i_3=i_1+i_2=i_1+2 i_1=3 i_1$

Now, we consider section $cc$,

Equating potentials, we get

$i_4 R=i_3\left(R+\frac{2}{3} R\right)$

$i_4=i_3\left(\frac{5}{3}\right)$

So, current, $i_5=i_4+i_3=\frac{5}{3} i_3+i_3=\frac{8}{3} i_3$

$=\frac{8}{3}\left(3 i_1\right)=8 i_1$

Similarly, across section $bb$,

$\Rightarrow \quad R i_6=\left(\frac{13}{8} R\right) i_5 \text { or } i_6=\frac{13}{8} i_5$

So, current,

$i_7=\frac{13}{8} i_5+i_5=\frac{21}{8} i_5=\frac{21}{8}\left(8 i_1\right)=21 i_1$

Now, for section $a a$, we have

$R i_8=i_7\left(\frac{13}{21}+1\right) R$

$R i_8=i_7\left(\frac{34}{21} R\right)$

$\Rightarrow i_8=i_7 \times \frac{34}{21}$

Hence, current $i$ is

$i=i_7+i_8$

$=i_7+\frac{34}{21} i_7=\frac{55}{21} i_7$

$=\frac{55}{21} \times 21 i_1$

$=55 i_1=55 \times 10^{-3} A$

$\left(\because i_1=1 \,mA , \text { given }\right)$

Total resistance across $P Q$ is

$R_{ eq }=\frac{34}{55} \,k \Omega$

$=\frac{34 \times 1000}{55} \Omega$

So, potential drop across, $P Q$

$=i R_{ eq }=55 \times 10^{-3} \times \frac{34}{55} \times 10^3 \,V$

$=34 \,V$