The figure given below shows an LCR series circuit with two...

From the source $v = V_0 \sin(300t)$, the angular frequency is $\omega = 300\text{ rad/s}$.
Capacitance $C = 100\ \mu\text{F} = 10^{-4}\text{ F}$, so $X_C = \dfrac{1}{\omega C}$.

Case 1 ($S_1$ closed, $S_2$ open):
Closing $S_1$ short-circuits $L_2$, leaving an LCR series circuit with $C$, $R$ and $L_1$. With phase difference $\phi_1 = 30^\circ$:
$\tan 30^\circ = \dfrac{\omega L_1 - X_C}{R}$
$\dfrac{1}{\sqrt{3}} = \dfrac{\omega L_1 - X_C}{R} \Rightarrow R = \sqrt{3}\,(\omega L_1 - X_C)$ ... (1)

Case 2 ($S_2$ closed, $S_1$ open):
Closing $S_2$ short-circuits $L_1$, leaving an LCR series circuit with $C$, $R$ and $L_2$. With phase difference $\phi_2 = 60^\circ$:
$\tan 60^\circ = \dfrac{\omega L_2 - X_C}{R}$
$\sqrt{3} = \dfrac{\omega L_2 - X_C}{R} \Rightarrow \omega L_2 - X_C = R\sqrt{3}$ ... (2)

Substituting $R$ from (1) into (2):
$\omega L_2 - X_C = \sqrt{3}\,(\omega L_1 - X_C) \cdot \sqrt{3}$
$\omega L_2 - X_C = 3\omega L_1 - 3X_C$
$2X_C = 3\omega L_1 - \omega L_2 = \omega(3L_1 - L_2)$

Therefore:
$3L_1 - L_2 = \dfrac{2X_C}{\omega} = \dfrac{2}{\omega^2 C}$

Substituting $\omega = 300\text{ rad/s}$ and $C = 10^{-4}\text{ F}$:
$\omega^2 C = (300)^2 \cdot 10^{-4} = 9 \times 10^4 \cdot 10^{-4} = 9$

$3L_1 - L_2 = \dfrac{2}{9}\text{ H}$

Hence, the correct option is $(2)\ \dfrac{2}{9}$.