For circuit $(a)$, the equivalent resistance is

$R_{ eq }=\frac{300 \times R}{300+R}+(100+200)$

$=\frac{300 R+300 R+90000}{(300+R)}$

The current through the circuit,

$I=\frac{V}{R_{ eq }}$

$=\frac{(300+R) \times 10}{(600 R+90000)}$

Using voltage division rule, voltage across $R$ for circuit $(a)$,

$V_a=\frac{(300+R) \times 10}{(600 R+90000)} \times \frac{300 R}{(300+R)}$

$=\frac{10 R}{2R+300}$

For circuit $(b)$, the equivalent resistance is

$R_{ eq }=\frac{(200+R) \times 300}{500+R}+100$

$=\frac{110000+400 R}{500+R}$

The current in circuit $(b)$ is

$I=\frac{V}{R_{ eq }}=\frac{(500+R) \times 10}{(110000+400 R)}$

Again, voltage across $R$ for circuit $(b)$,

$V_b=\frac{(500+R) \times 10}{(110000+400 R)} \times \frac{300 R}{(500+R)}$

$=\frac{30 R}{1100+4R}$

According to question,

$V_a=V_b$

$\frac{10 R}{(2 R+300)}-\frac{30 R}{(1100+4 R)}$

$1100+4 R=6 R+900$

$\Rightarrow R =100 \,\Omega$