Answer
$a.\quad 2.5\mathrm{m}\mathrm{V}$
$ b.\quad$ increases.
Work Step by Step
$V=IR,\qquad R=\displaystyle \rho\frac{L}{A},\ \ \ \ \rho=1.68\times 10^{-8}\Omega\cdot \mathrm{m}$
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$(\mathrm{a})$
$V=IR=I\displaystyle[ \rho\frac{L}{A}]=(32\mathrm{A})(1.68\times 10^{-8}\Omega\cdot \mathrm{m})\frac{0.060\mathrm{m}}{0.13\times 10^{-4}\mathrm{m}^{2}}=2.5\mathrm{m}\mathrm{V}$
2. (b)
From $R=\displaystyle \rho\frac{L}{A}$, we see that $R$ is proportional to $L$.
From $V=IR $, we see that $R$ is proportional to $V$
Thus $V$ is proportional to $L$.
If L increases, so does V.