Given circuits are Circuit $I$,

Clearly, both circuits are different af ter sections $A B$ and $C D$.

So, resistances of two circuits are same, if $R_{A B}=R_{C D}$

$\text { So, } \quad R+X=R+\frac{6 R(R+X)}{6 R+R+X}$

$\Rightarrow \quad X=\frac{6 R(R+X)}{7 R+X}$

$\Rightarrow 7 R X+X^2=6 R^2+6 R X$

$\Rightarrow X^2+R X=6 R^2=0$

From sridharacharya formula, we have

$\Rightarrow \quad X=\frac{-R^2 \pm \sqrt{R^2-4(1)\left(-6 R^2\right)}}{2(1)}$

$=\frac{-R \pm \sqrt{25 R^i}}{2}$

or $X=\frac{-R \pm 5 R}{2}$

$\therefore \quad X=2 R$