Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the wire is attached to the two ends of the battery directly, it has the same potential difference of the battery.
$$\Delta V_{\rm wire}=\Delta V_{\rm Battery}$$
We know that the electric field is given by
$$E=\dfrac{\Delta V}{L}$$
So, the electric field inside the wire is given by
$$E=\dfrac{\Delta V_{\rm wire}}{L_{\rm wire}}=\dfrac{1.5}{0.15}$$
$$E=\color{red}{\bf 10}\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
We know that the current density is given by
$$J=\sigma E $$
So,
$$J_{\rm wire}=\sigma_{\rm nichrome} E=(6.7\times 10^5)(10) $$
$$J_{\rm wire}=\color{red}{\bf 6.7 }\;\rm MA/m^2$$
$$\color{blue}{\bf [c]}$$
We know that the current density is given by
$$J=\sigma E=\dfrac{I}{A}$$
So,
$$A=\dfrac{I}{\sigma E}$$
where $A=\pi r^2=\pi (D/2)^2$ where $D$ is the diameter.
Hence,
$$D^2=\dfrac{4I}{\pi \sigma E}$$
$$D =\sqrt{\dfrac{4I}{\pi \sigma E}}$$
Plug the known;
$$D =\sqrt{\dfrac{4(2)}{(6.7\times 10^5)(10) }}$$
$$D=\color{red}{\bf 1.09}\;\rm mm$$
