$\mathrm{I}=\frac{2}{10+\mathrm{R}}$

The potential difference across the potentiometer wire is $\mathrm{V}=$ current $\times$ resistance

$=\frac{2}{10+\mathrm{R}} \times 10$

The length of the wire is $l=100 \mathrm{\,cm}$

So, the potential gradient along the wire is

$\mathrm{K}=\frac{\mathrm{V}}{\mathrm{t}}=\left(\frac{2}{10+\mathrm{R}}\right) \times \frac{10}{100}$          ...........$(i)$

The source of $\mathrm{emf}$ $10 \mathrm{\,mV}$ is balanced against a length of $40 \mathrm{\,cm}$ of the potentiometer wire

i.e. $10 \times 10^{-3}=\mathrm{k} \times 40$

or $10 \times 10^{-3}=\frac{2}{(10+\mathrm{R})} \times \frac{40}{10}$      (Using $(i)$)

or $\quad \mathrm{R}=790\, \Omega.$