b
Current through the wire of $5$ $\mathrm{ohm}$
$=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}}=\frac{3}{1+5}=0.5 \mathrm{\,A}$
Potential cifference actioss the wire of $5 \Omega=0.5 \times 5=2.5 \mathrm{\,V}$
Length of wire $=1 \mathrm{\,m}$
$\therefore $ Potential gradient $=2.5 \mathrm{\,V} / \mathrm{m}$