Step 1: Turn the sensitivities into basic quantities.
Current sensitivity is $10$ divisions per mA, so one division needs \[ I_{div} = \frac{1\,\text{mA}}{10} = 10^{-4}\ \text{A}. \] Voltage sensitivity is $2$ divisions per mV, so one division needs $V_{div} = \frac{1\,\text{mV}}{2} = 0.5\times10^{-3}\,\text{V}$.
Step 2: Get the galvanometer resistance.
Since voltage sensitivity equals current sensitivity divided by $G$, we have \[ G = \frac{V_{div}}{I_{div}} = \frac{0.5\times10^{-3}}{10^{-4}} = 5\ \Omega. \]
Step 3: Find the full-scale values we want.
We want each division to read $1\,\text{V}$, and there are $150$ divisions, so full scale is $V = 150\,\text{V}$. The full-scale current stays $I_g = 150 \times 10^{-4} = 0.015\,\text{A}$.
Step 4: Use the voltmeter conversion formula.
A series resistance $R$ turns the galvanometer into a voltmeter via $V = I_g (R + G)$, so \[ R + G = \frac{V}{I_g} = \frac{150}{0.015} = 10000\ \Omega. \]
Step 5: Subtract the meter resistance.
\[ R = 10000 - 5 = 9995\ \Omega. \]
Step 6: Conclusion.
The resistance to be put in series is $9995\,\Omega$. \[ \boxed{9995\ \Omega} \]