1. Fundamental frequency of a closed organ pipe

For a closed organ pipe, $f_1=\frac{v}{4L}$ where $L$ is the air column length.

2. Initial length of air column

Given initial fundamental frequency: $f_1=30\,\text{Hz}$ $L_1=\frac{v}{4f_1}=\frac{330}{4\times 30}=\frac{330}{120}=2.75\,\text{m}$

3. Final length of air column after pouring water

New fundamental frequency: $f_2=110\,\text{Hz}$ So new air column length is $L_2=\frac{v}{4f_2}=\frac{330}{4\times 110}=\frac{330}{440}=0.75\,\text{m}$

4. Height of water poured

The water reduces the air column length, so height of water added is $h=L_1-L_2=2.75-0.75=2.0\,\text{m}$

5. Volume of water poured

Cross-sectional area: $A=2\,\text{cm}^2=2\times 10^{-4}\,\text{m}^2$

Thus volume of water is $V=Ah=(2\times 10^{-4})(2.0)=4\times 10^{-4}\,\text{m}^3$

Convert to $\text{cm}^3$: $1\,\text{m}^3=10^6\,\text{cm}^3$ $V=4\times 10^{-4}\times 10^6=400\,\text{cm}^3$

6. Mass of water

Since density of water is $1\,\text{g/cm}^3$, $m=400\,\text{g}$

7. Final answer

The amount of water poured is $\boxed{400}$ (g)