#### Answer

(a) $E = 10~V/m$
(b) $J = 9.1\times 10^6~A/m^2$
(c) The diameter of the wire is $0.53~mm$

#### Work Step by Step

(a) We can find the electric field:
$E = \frac{\Delta V}{d}$
$E = \frac{1.5~V}{0.15~m}$
$E = 10~V/m$
(b) We can find the current density:
$J = \sigma~E$
$J = (9.1\times 10^5~S/m)(10~V/m)$
$J = 9.1\times 10^6~A/m^2$
(c) We can find the radius:
$I = J~A$
$I = J~\pi~r^2$
$r^2 = \frac{I}{J~\pi}$
$r = \sqrt{\frac{I}{J~\pi}}$
$r = \sqrt{\frac{2.0~A}{(9.1\times 10^6~A/m^2)~(\pi)}}$
$r = 0.264\times 10^{-3}~m$
$r = 0.264~mm$
Since the diameter is double the radius, the diameter of the wire is $0.53~mm$