(b) $4\pi\varepsilon_{0}r = \frac{\varepsilon_{0}A}{d} \Rightarrow d = \frac{A}{4\pi r} = \frac{\pi\left( 20 \times 10^{- 3} \right)^{2}}{4\pi \times 1} = 0.1\ mm$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.