The resistance of the circuit between the points A and B of figure shown is $b\Omega$. The the value of '$b$' is

For finding R_AB we will convert the delta CDE of Fig. into its equivalent star as shown in Figure
R_CS=8 x 4/18 = 16/9$\Omega$ ; R_ES=8 x 6/18 =24/9$\mathrm{\Omega }$: R_DS=6 x 4/18=12/9$\mathrm{\Omega }$

The two parallel resistances between S and B can be reduced to a single resistance of 35 / 9 $\mathrm{\Omega }$

As seen from figure, R_AB=4+(76/9)+(35/9)=87/9$\mathrm{\Omega }$

Alternative method: The circuit has antisymmetric WheatStone bridge. Let $i_1$   be the current through the $8\Omega$  and $i_2$ be the current through $4\Omega$  resistance of the bridge. Therefore, current through $4\Omega$  resistance that is outside the bridge is  $\left(i_1+i_2\right)$

Now using KVL, we can write equations,
Loop  $ABEDA$,

$-8i_1-6\left(i_1-i_2\right)+4i_2=0$, 

$7i_1=5i_2$                      …(1)
Loop $ABCVA$, 
 $-4\left(i_1+i_2\right)-8i_1-4i_2+V=0$

$12i_1+8i_2=V$            …(2)
For the entire circuit,
$V=\left(i_1+i_2\right)R_{\text{eff}}$        …(3)
Using first two equations, get $i_1$ and $i_2$  in terms of $V$
 $i_1=\frac{5V}{116};i_2=\frac{7V}{116}$
Using above in (3),

$V=\left(\frac{5V}{116}+\frac{7V}{116}\right)R_{\text{eff}}$

$V=\frac{12V}{116}\times R_{\text{eff}}$
 
 $V=\frac{12V}{116}\times R_{\text{eff}}\Rightarrow R_{\text{eff}}=\frac{116}{12}=9.67\Omega$