The increase in pressure required to decrease the volume of a water sample by  $$0.2%$$  is  $$P \times10^5\,Nm^{-2}.$$ Bulk modulus of water is  $$2.15\times10^9\,Nm^{-2}.$$  The value of  $$P$$  is  $$\underline{\hspace{2cm}}.$$

We need to find the pressure increase required to decrease the volume of a water sample by 0.2%.

We know that the bulk modulus $$B$$ is defined as $$B = -\frac{\Delta P}{\Delta V / V}$$, where $$\Delta P$$ is the change in pressure and $$\Delta V / V$$ is the fractional change in volume; the negative sign indicates that an increase in pressure causes a decrease in volume. Hence, the magnitude of the pressure change for a given compression is given by $$\Delta P = B \times \frac{|\Delta V|}{V}$$.

Since the fractional volume change is $$\frac{|\Delta V|}{V} = 0.2\% = \frac{0.2}{100} = 0.002$$ and the bulk modulus is $$B = 2.15 \times 10^9 \, \text{N m}^{-2}$$, substituting these values into the formula yields $$\Delta P = 2.15 \times 10^9 \times 0.002$$.

Performing the multiplication gives $$\Delta P = 2.15 \times 0.002 \times 10^9 = 0.0043 \times 10^9 = 4.3 \times 10^6 \, \text{N m}^{-2}$$.

Expressing this in the form $$P \times 10^5 \, \text{N m}^{-2}$$, we obtain $$4.3 \times 10^6 = 43 \times 10^5 \, \text{N m}^{-2}$$, so that $$P = 43$$.

The answer is 43.