The driver cell of a potentiometer has an emf of 2V and negligible internal resistance.  The potentiometer wire has a resistance of 5$\Omega$  and is 1m long.  The resistance which must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is

1995 $\Omega$ 

In order to have a potential drop of 5 mV , across a wire of resistance 5 W, the current flowing in the wire should be:

$ I = \frac{5\times 10^{-3}}{5} = 10^{-3}A$

If R is the resistance to be connected in series with the wire, then $\frac{2}{R+5} = 10^{-3}A$

which gives $R = 1995 \Omega$