Answer
${\bf 9.11}\;\rm A$
Work Step by Step
We know, according to Ohm's law, that
$$\Delta V_{\rm tube}=IR$$
and since the battery is connected directly to the hollow nichrome tube, the hollow nichrome tube must have the same potential difference.
Thus,
$$\Delta V_{\rm battery}=IR$$
where $R=\rho L/A$
$$\Delta V_{\rm battery}=I\dfrac{\rho_{\rm nichrome }L}{A}$$
where $A$ is the cross-sectional area of the tube that faces the current which is given by $A=A_{out}-A_{in}=\pi (r^2_{\rm out}-r^2_{\rm in})$
$$\Delta V_{\rm battery}=I\dfrac{\rho_{\rm nichrome }L}{\pi (r^2_{\rm out}-r^2_{\rm in})}$$
Solving for $I$;
$$I=\dfrac{\pi (r^2_{\rm out}-r^2_{\rm in}) \Delta V_{\rm battery}} {\rho_{\rm nichrome }L}$$
Plug the known;
$$I=\dfrac{\pi ([1.5\times 10^{-3}]^2 -[1.4\times 10^{-3}]^2 ) (3)} {(1.5\times 10^{-6})(0.20)}$$
$$I=\color{red}{\bf 9.11}\;\rm A$$
