The pressure in air bubble just below the water surface is (Surface tension of liquid is 72dyne cm^-1 and atmospheric pressure = $1.0\times {10}^5{\mathrm{Nm}}^{-2}$, Radius of bubble = 0.1mm)

Given,

Radius of bubble = 0.1mm

surface tension = $\sigma =72dyne/cm=72\times {10}^{-3}N/m$

As bubble is just below the water surface so there will be only one transition of medium that is water to air,

Thus the pressure difference is 

$∆\mathrm{P}=\frac{2\mathrm{S}}{\mathrm{r}}$

$\mathrm{P}-{\mathrm{P}}_0=\frac{2(72\times {10}^{-3})}{0.1\times {10}^{-3}}$

$\Rightarrow \mathrm{P}=1440+1\times {10}^5=(1+0.0144)\times {10}^5=1.0144\times {10}^5{\mathrm{Nm}}^{-2}$