Explanation

(a) Self inductance of a solenoid
$L = \frac{\mu_{0}N^{2}A}{l} = \frac{\mu_{0}N^{2}\pi r^{2}}{l}$
Where $l$ is the length of the solenoid, $N$ is the total number of turns of the solenoid and $A$ is the area of cross-section of the solenoid
$\therefore\frac{L_{1}}{L_{2}} = \left( \frac{N_{1}}{N_{2}} \right)^{2}\left( \frac{r_{1}}{r_{2}} \right)^{2}\left( \frac{l_{2}}{l_{1}} \right)$
Here, $N_{1} = N_{2},\frac{l_{1}}{l_{2}} = \frac{1}{2},\frac{r_{1}}{r_{2}} = \frac{1}{2}$
$\therefore\frac{L_{1}}{L_{2}} = \left( \frac{1}{2} \right)^{2}\left( \frac{2}{1} \right) = \frac{1}{2}$