Answer
$ \rho=7.08\times 10^{-8}\Omega m$
Work Step by Step
We know that
$R=\frac{\rho L}{A}$
We plug in the known values to obtain:
$R=\frac{\rho\times 6.9}{\pi(0.165\times 10^{-3})^2}$
We also know that
$V=IR$
We plug in the known values to obtain:
$12=2.1\times \frac{\rho\times 6.9}{\pi(0.165\times 10^{-3})^2}$
$\implies \rho=\frac{12\times \pi (0.165\times 10^{-3})^2}{2.1\times 6.9}$
$\implies \rho=7.08\times 10^{-8}\Omega m$
