In the given circuit below inductance values of L_1 , L_2 and L_3...

Let the inductance of each inductor be $L$, so $L_1 = L_2 = L_3 = L$.

Let the total current flowing through the circuit be $I$. This current flows entirely through $L_1$.

Since $L_2$ and $L_3$ are connected in parallel and have equal inductances, the current $I$ divides equally between them. Therefore, the current through $L_2$ is $\dfrac{I}{2}$ and the current through $L_3$ is $\dfrac{I}{2}$.

The magnetic energy stored in an inductor is given by $U = \dfrac{1}{2} L i^2$.

The energy stored in the inductor $L_2$ is:
$U_l = \dfrac{1}{2} L \left(\dfrac{I}{2}\right)^2 = \dfrac{1}{8} L I^2$

The equivalent inductance of the entire circuit is:
$L_{eq} = L_1 + \dfrac{L_2 L_3}{L_2 + L_3} = L + \dfrac{L \cdot L}{L + L} = L + \dfrac{L}{2} = \dfrac{3L}{2}$

The total magnetic energy stored in the entire circuit is:
$U_t = \dfrac{1}{2} L_{eq} I^2 = \dfrac{1}{2} \left(\dfrac{3L}{2}\right) I^2 = \dfrac{3}{4} L I^2$

The ratio of the total energy to the energy stored in $L_2$ is:
$\dfrac{U_t}{U_l} = \dfrac{\dfrac{3}{4} L I^2}{\dfrac{1}{8} L I^2} = \dfrac{3}{4} \times 8 = 6$

Answer: $6$