A uniform metallic wire carries a current 2   A , when 3 . 4   V battery is connected across it.…

Given here, $I=2 \mathrm{A}, \mathrm{V}=3.4 \mathrm{V}, m=8.92\times {10}^{-3} \mathrm{kg},$

$d=8.92\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3} \mathrm{and} \rho =1.7\times {10}^{-8} \mathrm{\Omega }-\mathrm{m}$

From Ohm's law, resistance is $\mathrm{R}=\frac{V}{I}=\frac{3.4}{2}=1.7 \mathrm{\Omega }$

As resistance of wire is given by $R=\frac{\rho l}{A}$, where, $A$ is area and $l$ is length of wire.

 $\Rightarrow 1.7=\frac{\rho l}{A}$ $\Rightarrow l=\frac{1.7A}{\rho }$

Now, volume$=\frac{m}{d}=\frac{8.92\times {10}^{-3}}{8.92\times {10}^3}={10}^{-6} {\mathrm{m}}^3$

Thus, 

$$\begin{gathered} l^2=\frac{1.7A}{\rho }\times l \\ =\frac{1.7\left({\displaystyle \frac{\text{volume}}{l}}\right)}{\rho }\times l \\ =\frac{1.7}{\rho }\left({10}^{-6}\right) \\ =\frac{1.7}{1.7\times {10}^{-8}}\times {10}^{-6} \end{gathered}$$

Hence, $l=10 \mathrm{m}$