A galvanometer has a resistance of 100 Ω and shows full deflection for 2.0 mA…

Step 1. The shunt. $S = \dfrac{I_g G}{I - I_g}$ $= \dfrac{2.0 \times 10^{-3}\ \text{A} \times 100\ \Omega}{3.0\ \text{A} - 2.0 \times 10^{-3}\ \text{A}}$ $= 0.0667\ \Omega = 66.7\ \text{m}\Omega$.

Step 2. The ammeter. Its resistance is $\dfrac{GS}{G + S} = \dfrac{100 \times 0.0667}{100.0667}\ \Omega$ $= 0.0667\ \Omega$, about $66.7\ \text{m}\Omega$.

Step 3. The voltmeter. The total resistance is $\dfrac{V}{I_g} = \dfrac{6.0\ \text{V}}{2.0 \times 10^{-3}\ \text{A}}$ $= 3000\ \Omega$. The series resistor is $R = 3000\ \Omega - 100\ \Omega$ $= 2900\ \Omega$.