A galvanometer has a coil resistance of 75 Ω and a full-scale current of 0.40 mA…

Step 1. The $12$ V range. $R = \dfrac{V}{I_g} - G = \dfrac{12\ \text{V}}{0.40 \times 10^{-3}\ \text{A}} - 75\ \Omega = 30000\ \Omega - 75\ \Omega = 29925\ \Omega$, that is $29.9\ \text{k}\Omega$.

Step 2. The $6.0$ V range. $R = \dfrac{6.0\ \text{V}}{0.40 \times 10^{-3}\ \text{A}} - 75\ \Omega = 15000\ \Omega - 75\ \Omega = 14925\ \Omega$, that is $14.9\ \text{k}\Omega$.

Step 3. The resistance of the $12$ V voltmeter. $R_V = G + R = 75\ \Omega + 29925\ \Omega = 30000\ \Omega = 30.0\ \text{k}\Omega$.

Step 4. Ohms for each volt. $\dfrac{30000\ \Omega}{12\ \text{V}} = 2500\ \Omega/\text{V}$. This is just $\dfrac{1}{I_g}$, the same for every range of this coil.