Answer:
Option C
Explanation:
According to question , magnetic
permeability of the rod is given by , $\mu_{r}=\frac{B}{B_{0}}$
where, B= magnetic field without iron rod
B0 = magnetic field after insertion of iron rod
so, $\mu_{r}=\frac{B}{2000Am^{-1}}$
$\left( \because B.A=\phi\Rightarrow B=\frac{\phi}{A}=\frac{6 \times 10 ^{-4}Wb}{3\times 10^{-4}m^{2}}=2 Wbm^{-2}\right)$
So, $\mu_{r}=\frac{2}{2000}=10^{-3}$