a
Given : Current through the galvanometer, $i_{g}=5 \times 10^{-3} \,A$
Galvanometer resistance, $G=15\, \Omega$
Let resistance $R$ to be put in series with the galvanometer to convert it into a voltmeter.
$V=i_{g}(R+G)$
$10= 5 \times 10^{-3}(R+15) $
$ \therefore \quad R=2000-15=1985 $
$=1.985 \times 10^{3} \,\Omega $