Answer
$0.61\;\rm A$
Work Step by Step
According to Ohm's law
$$V_B=IR$$
and our circuit here consists of the wire and the battery.
So,
$$I=\dfrac{V_B}{R}\tag 1$$
Now we need to find $R$ where we know that
$$R=\dfrac{\rho L }{A }$$
Hence,
$$dR=\dfrac{\rho dL }{A }$$
replacing $dL$ by $dx$;
$$dR=\dfrac{\rho dx }{A }$$
Integrating both sides,
$$\int_0^RdR=\int_0^L\dfrac{\rho dx }{A }$$
$$ R=\dfrac{1 }{A }\int_0^L\rho dx$$
where $\rho =(2.5\times 10^{-6})\left[1+x^2\right]$
$$ R=\dfrac{4(2.5\times 10^{-6}) }{\pi D^2 }\int_0^L\left[1+x^2\right]dx$$
$$ R=\dfrac{4(2.5\times 10^{-6}) }{\pi D^2 }\left[x+\frac{1}{3}x^3\right]_0^L$$
$$ R=\dfrac{4(2.5\times 10^{-6}) }{\pi D^2 }\left[L+\frac{1}{3}L^3\right] $$
Plug the given;
$$ R=\dfrac{4(2.5\times 10^{-6}) }{\pi (1\times 10^{-3})^2 }\left[2+\frac{2^3}{3}\right] =\bf 14.85\;\rm\Omega$$
Thus, from (1),
$$I=\dfrac{V_B}{R}=\dfrac{9}{14.85}$$
$$I=\color{red}{\bf 0.61}\;\rm A$$
