Find the compressibility of water (B = 2.2 × 10⁹ Pa), of glycerine (B = 4.5 × 10⁹ Pa), and of air…

Step 1. Water. $\dfrac{1}{2.2 \times 10^{9}\ \text{Pa}} = 4.5 \times 10^{-10}\ \text{Pa}^{-1}$.

Step 2. Glycerine. $\dfrac{1}{4.5 \times 10^{9}\ \text{Pa}} = 2.2 \times 10^{-10}\ \text{Pa}^{-1}$.

Step 3. Air. $B = p = 1.0 \times 10^{5}\ \text{Pa}$, so $k = 1.0 \times 10^{-5}\ \text{Pa}^{-1}$.

Step 4. Ratio. $\dfrac{1.0 \times 10^{-5}\ \text{Pa}^{-1}}{4.5 \times 10^{-10}\ \text{Pa}^{-1}} = 2.2 \times 10^{4}$.