(d) Given $2\pi R = 2 \Rightarrow R = \frac{1}{\pi}$ For sphere $C = 4\pi\varepsilon_{0}KR$ $\Rightarrow C = \frac{1}{9 \times 10^{9}} \times \frac{1}{\pi} \times 80\ \ \ \ \ $[For water $K = 80$] $\Rightarrow C = 2828.28\ PF$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.