b
Let additional resistance is $\mathrm{R}$
then $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{R}_{\mathrm{wire}}}=\frac{2}{\mathrm{R}+3}$
$\phi = \frac{{{V_{{\rm{wire }}}}}}{\ell } = \frac{{I{R_{{\rm{wire }}}}}}{\ell }$
$\frac{10^{-3}}{10^{-2}}=\left(\frac{2}{\mathrm{R}+3}\right) \frac{3}{(1)}$ solving this, we get
$\mathrm{R}=57 \,\Omega$