1. Given data

• Radius of air bubble: $r = 1.0\text{ mm} = 1.0 \times 10^{-3}\text{ m}$
• Depth below liquid surface: $h = 20\text{ cm} = 0.20\text{ m}$
• Surface tension: $T = 0.095\text{ N/m}$
• Density of liquid: $\rho = 10^3\text{ kg/m}^3$
• Acceleration due to gravity: $g = 10\text{ m/s}^2$

We need:

$P_{\text{inside}} - P_{\text{atm}}$

2. Pressure outside the bubble

At depth $h$ in a liquid, the pressure is:

$P_{\text{outside}} = P_{\text{atm}} + \rho g h$

So,

$\rho g h = (10^3)(10)(0.20) = 2000\text{ N/m}^2$

Hence,

$P_{\text{outside}} = P_{\text{atm}} + 2000$

3. Excess pressure inside an air bubble in a liquid

For an air bubble in a liquid, there is only one liquid surface, so excess pressure is:

$P_{\text{inside}} - P_{\text{outside}} = \frac{2T}{r}$

Substitute values:

$\frac{2T}{r} = \frac{2(0.095)}{1.0 \times 10^{-3}} = \frac{0.19}{10^{-3}} = 190\text{ N/m}^2$

4. Pressure inside bubble relative to atmospheric pressure

$P_{\text{inside}} - P_{\text{atm}} = (P_{\text{outside}} - P_{\text{atm}}) + (P_{\text{inside}} - P_{\text{outside}})$

$= 2000 + 190 = 2190\text{ N/m}^2$

5. Final answer

$\boxed{2190\text{ N/m}^2}$