\( R = \frac{1}{a} p \implies p = \frac{R.a}{1} = \frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2 \\
\text{Therefore } p = \frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4} \\
= 0.3925 \times 10^{-7} \text{ OR } -3.930 \times 10 ^{-8}Ωm \)