a
Required potential gradient \(=1 \,\mathrm{mV} \,\mathrm{cm}^{-1}\)

\(=\frac{1}{10}\, \mathrm{Vm}^{-1}\) 

Length of potentiometer wire, \(l=4\, \mathrm{m}\)

So potential difference across potentiometer wire

\(=\frac{1}{10} \times 4=0.4\, \mathrm{V}\)     ....\((i)\)

In the circuit, potential difference across \(8 \,\Omega\)

\(=I \times 8=\frac{2}{8+R} \times 8\)       ....\((ii)\)

Using equation \((i)\) and \((ii),\) we get,

\(0.4=\frac{2}{8+R} \times 8\) \(\frac{4}{10}=\frac{16}{8+R}, \,\,8+R=40\)

\(\therefore \quad R=32 \,\Omega\)