d
$5=\lambda \ell$
where $\lambda$ is potential gradient and $L$ is total length of wire.
$5=\frac{\Delta V}{L} \ell$
$\Delta \mathrm{V}=\frac{5 \times \mathrm{L}}{\ell}=5 \times \frac{12}{10}=6 \mathrm{V}=60 \;\mathrm{mA} \times \mathrm{R}$
$\mathrm{R}=100 \;\Omega$