Solution :
A galvanometer can be converted into a voltmeter.
When a suitable resistance of R is connected in series with the galvanometer.
$R= \large\frac{v}{I_g}$$-a=\large\frac{7.5}{2.5 \times 10^{-3}}$$-12$
$\qquad= 2988\;\Omega$
Total resistance of voltmeter $=R+G$
$\qquad= 2988+12$
$\qquad= 3000\;\Omega$
Answer : $3000 \;\Omega $