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