The angular frequency of alternating current in a L-C-R circuit

Current through $60 \Omega$ resistance $=\frac{15}{60}=\frac{1}{4} \mathrm{~A}$

thus capacitor current $=\frac{1}{4} \mathrm{~A}$

$\because \quad \mathrm{V}_{\mathrm{C}}=\mathrm{I} \mathrm{X}_{\mathrm{C}}$

$10=\frac{1}{4} \times \frac{1}{\omega C}$

$\therefore \mathrm{C}=\frac{1}{40 \omega}=\frac{1}{4000}=250 \mu \mathrm{F}$

Now,

current through $40 \Omega$ resistance $=\frac{20}{40}=\frac{1}{2} \mathrm{~A}$

thus current through inductor $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \mathrm{~A}$

$\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{1}{4} \times \omega \mathrm{L}$

$20=\frac{1}{4} \times 100 \times \mathrm{L}$

$\Rightarrow \mathrm{L}=0.8 \mathrm{H}$