Answer:
Option D
Explanation:
According to the question
$v_{d}=2.5\times 10^{-4}m/s$
$\Rightarrow$ $n=8\times 10^{28}/m^{3}$
we know that $J=nev_{d}$ or $I=nev_{d}A$
where, symbols have their usual meaning
$\Rightarrow$ $\frac{V}{R}=nev_{d}A$
or $\frac{V}{\frac{\rho L}{A}}=nev_{d}A$
or $\frac{V}{{\rho L}}=nev_{d}$
or $\rho =\frac{V}{nev_{d}L}$
= $\frac{5}{8\times 10^{28}\times1.6\times 10^{-19}\times2.5 \times 10^{-4}\times0.1}$
or $\rho$ =$1.6\times 10^{-5}$ Ωm