$$ P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r} $$

Given that the pressure inside the bubble is $$P_{\text{atm}} + 2100\,\text{N/m}^2$$ and the liquid pressure at a depth of 20 cm is

$$ P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2, $$

the pressure difference due solely to surface tension is

$$ \Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2. $$

The Laplace pressure difference for a bubble (which has one interface) is given by

$$ \Delta P = \frac{2S}{r}. $$

The given bubble radius is 0.1 cm, which in SI units is

$$ r = 0.1\,\text{cm} = 0.001\,\text{m}. $$

Substituting the known values into the Laplace equation gives

$$ \frac{2S}{0.001} = 100. $$

Solving for $$S$$:

$$ 2S = 100 \times 0.001 = 0.1, $$

$$ S = \frac{0.1}{2} = 0.05\,\text{N/m}. $$

Thus, the surface tension of the liquid is

$$ \boxed{0.05\,\text{N/m}}. $$