first we distribute current in circuit given as

Current distribution must follows Kirchhoff's junction rule.

Now, from closed loops marked $1, 2, 3$ and $4$ , we have following set of equations by application of Kirchhoff's loop rule,

$I_1=I_2+I_3 \quad \dots(i)$

$I_3=I_2+I_4 \quad \dots(ii)$

$I_4=I_2-I_4+I_5$

$\Rightarrow 2 I_4=I_2+I_5 \quad \dots(iii)$

$I_5=2\left(I_2-I_4-I_5\right)$

$\Rightarrow I_5=2 I_2-2 I_4-2 I_5 \quad \dots(iv)$

$3 I_5=2 I_2-2 I_4 \quad \dots(v)$

From Eqs. $(iii)$ and $(v)$, we have

$3 I_5=2 I_2-\left(I_2+I_5\right)$

$\Rightarrow 4 I_5=I_2 \quad \dots(vi)$

From Eqs. $(iii)$ and $(vi)$, we have

$2 I_4=4 I_5+I_5 \Rightarrow I_4=\frac{5}{2} I_5 \quad \dots(vii)$

From Eqs. $(ii), (vi)$ and $(vii)$, we have

$I_3=4 I_5+\frac{5}{2} I_5=\frac{13}{2} I_5 \quad \dots(viii)$

Now, marked currents $I$ and $I^{\prime}$ in the given circuit are

$I^{\prime}=\left(I_2-I_4-I_5\right)=\left(4 I_5-\frac{5}{2} I_5-I_5\right)$

$=\left(\frac{8-5-2}{2}\right) I_5=\frac{I_5}{2} \quad \dots(ix)$

And $I=I_2=4 I_5$

Hence, ratio of $I / I^{\prime}=\left(4 I_5\right) /\left(I_5 / 2\right)=8$