The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $$0.02\%$$ is _____ m. (Take density of sea water $$= 10^3 \text{ kg m}^{-3}$$, Bulk modulus of rubber $$= 9 \times 10^8 \text{ N m}^{-2}$$, and $$g = 10 \text{ m s}^{-2}$$)

$$\Delta P = \rho g h$$, $$\frac{\Delta V}{V} = \frac{\Delta P}{B}$$.

$$0.02\% = 0.0002 = \frac{\rho g h}{B} = \frac{10^3 \times 10 \times h}{9 \times 10^8}$$.

$$h = \frac{0.0002 \times 9 \times 10^8}{10^4} = \frac{1.8 \times 10^5}{10^4} = 18$$ m.

The answer is $$\boxed{18}$$.