Explanation:
D Given,
Resistance of galvanometer coil, $\mathrm{G}=12 \Omega$
Current for which there is full scale deflection $\mathrm{i}=3 \mathrm{~mA}=3 \times 10^{-3} \mathrm{~A}$.
A range of voltmeter is $\mathrm{O}$, which needs to be converted to $18 \mathrm{~V}$.
So,
$\mathrm{V}=18 \mathrm{~V}$
Let as a resistor of resistance $\mathrm{R}$ be connected in series with the galvanometer to convert it into a voltmeter. Then,

Then, we can write resistance of series resistor is -
$\mathrm{R} =\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{R}_{\mathrm{G}}$
$\mathrm{R} =\frac{18}{3 \times 10^{-3}}-1 \Omega$
$\mathrm{R} =5988 \Omega$
Hence, converted into a voltmeter of range $\mathrm{O}$ to $18 \mathrm{~V}$, add resistance of value $5988 \Omega$ is connected in series.