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

Current through $60Ω$ resistance $= \dfrac{15}{60} = \dfrac{1}{4}A$

thus capacitor current $= \dfrac{1}{4}A$

$∵ {V}_{C} = I{X}_{C}$

$10 = \dfrac{1}{4} \times \dfrac{1}{\omegaC}$

$∴ C = \dfrac{1}{40\omega} = \dfrac{1}{4000} = 250\muF$

Now,

current through $40Ω$ resistance $= \dfrac{20}{40} = \dfrac{1}{2}A$

thus current through inductor $= \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}A$

${V}_{L} = I{X}_{L} = \dfrac{1}{4} \times \omegaL$

$20 = \dfrac{1}{4} \times 100 \times L$

$⇒ L = 0 . 8H$