A 10 m  long wire of resistance $20\Omega$ is connected in series with battery of e.m.f. 3V (negligible internal resistance) and a resistance of $10 \Omega$. The potential gradient along the wire in volt per metre is 

0.1

$\text{Current flowing through the wire is } I = \frac{V}{R+r} = \frac{3V}{20\Omega} = 0.1A$

$ \text{Voltage across the wire is } V'= 20\times 0.1 = 2V$

$\text{Voltage Gradient } = V/l = \frac{2}{10} = 0.2V/m$