A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA.

Resistance of the galvanometer coil, G = 15 Ω

Current for which the galvanometer shows full scale deflection,

$I_{\mathrm{g}}=4 \mathrm{~mA}=4 \times 10^{-3} \mathrm{~A}$

Range of the ammeter is 0, which needs to be converted to 6 A.

$\therefore$ Current, $I=6 \mathrm{~A}$

A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:

$S=\frac{I_{g} G}{I-I_{g}}$

$=\frac{4 \times 10^{-3} \times 15}{6-4 \times 10^{-3}}$

$S=\frac{6 \times 10^{-2}}{6-0.004}=\frac{0.06}{5.996}$

$\approx 0.01 \Omega=10 \mathrm{~m} \Omega$

Hence, a $10 \mathrm{~m} \Omega$ shunt resistor is to be connected in parallel with the galvanometer.