Hello
1) First of all, since we know the radius of the wire ([tex]r=1.2~mm=0.0012~m[/tex]), we can calculate its cross-sectional area
[tex]A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2[/tex]
2) Then, we can calculate the current density J inside the wire. Since we know the current, [tex]I=3~A[/tex], and the area calculated at the previous step, we have
[tex]J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2[/tex]
3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity [tex]\sigma=3.6\cdot10^7~ \frac{A}{Vm} [/tex] of the aluminium, the electric field is given by
[tex]E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m[/tex]
