All resistances in figure are 1   Ω each. The value of current ' I ' is a 5   A . The…

Each pair on the right side of the circuit is in parallel connection, hence it can be replaced by a resistance of $\frac{r}{2}$ as shown.

Now, $r \& \frac{r}{2}$ in each branch on the right side is in series, hence combined resistance value will be, $\frac{3r}{2}$. Upper branch having resistance $\frac{3r}{2}$ is in parallel with another $\frac{3r}{2}$, hence it can be written as $\frac{3r}{4}$.

Further, simplified circuit will have net resistance of $\frac{7r}{8}$.

Therefore, $R_{\mathrm{eq}}=\frac{7r}{8}+r=\frac{7}{8}+1=\frac{15}{8} \mathrm{\Omega }$

Now, current $I=\frac{3}{{\displaystyle \frac{15}{8}}}=\frac{8}{5} \mathrm{A}$

Hence, value of $a=8$.