All resistances in figure are 1Ω each. The value of current ' I '...

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

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

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

Therefore, ${R}_{eq} = \dfrac{7r}{8} + r = \dfrac{7}{8} + 1 = \dfrac{15}{8}Ω$

Now, current $I = \dfrac{3}{\dfrac{15}{8}} = \dfrac{8}{5}A$

Hence, value of $a = 8$.