Let depth of bubble below the liquid surface, $$h = ?$$
As we know, $${\rho _{{\text{Excess}}}} = h\rho g + \frac{{2s}}{r}$$
$$\eqalign{ & \Rightarrow 1100 = h \times {10^3} \times 9.8 + \frac{{2 \times 6 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}} \cr & \Rightarrow 1100 = 9800\,h + 120 \Rightarrow 9800\,h = 1100 - 120 \cr & \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m \cr} $$

$$\eqalign{ & W = mg = \rho \left( {\pi {r^2}h} \right)g \cr & {\text{or}}\,\,W \simeq \rho \left( {\pi {r^2}} \right) \times \left( {\frac{{2T}}{{\rho rg}}} \right)g\,\,{\text{or}}\,\,W = 2\pi rT \cr & \therefore T = \frac{W}{{2\pi r}} = \frac{{6.28 \times {{10}^4}}}{{2\pi \times 2 \times {{10}^{ - 3}}}} = 5 \times {10^{ - 2}}\,N/m. \cr} $$