Correct Answer - Option 1 : 10 − 6 ohm
\(R\; = \;\frac{{\rho l}}{A}\; \Rightarrow \;50\; \times \;{10^{\; - \;8}}\; \times \;\frac{{50\; \times \;{{10}^{\; - \;2}}}}{{{{\left( {50\; \times \;{{10}^{\; - \;2}}} \right)}^2}}}\; = \;{10^{\; - \;6}}\;{\rm{\Omega }}\)