(c) From $Z_{0} = \sqrt{\frac{L}{C}}$ , $C = \frac{L}{Z_{0}^{2}} = \frac{0.4 \times 10^{- 3}}{160 \times 160}$ $= {15.62 \times 10}^{- 19}F = 15.62\ nF$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.