Solution :
A galvanometer can be coverted into am ammeter if a shunt resistance is connected in parallel in parallel to the galvanometer .
$S= \large\frac{I_g G}{I-I_g} =\frac{(2.5 \times 10^{-3}) \times 12}{7.5 -2.5 \times 10^{-3}}$
$\qquad= 4 \times 10^{-3} \Omega$
Effective resistance R of the amoents is
$R= \large\frac{GS}{G+S} =\frac{12 \times 4 \times 10^{-3}}{12+ 4 \times 10^{-3}}$
$\qquad= 4 \times 10^{-3}\;\Omega $
Answer : $ 4 \times 10^{-3}\;\Omega $