A wire forming a loop is submerged into a soap solution and...

The length of the thread loop is given by $L = 2\pi R = 6.28 \text{ cm} = 6.28 \times 10^{-2} \text{ m}$.

The radius of the circular loop is $R = \dfrac{6.28 \times 10^{-2}}{2 \times 3.14} = 10^{-2} \text{ m}$.

When the film inside the loop is pricked, the outer soap film pulls the thread radially outward. Since a soap film has two free surfaces, the outward force per unit length is $2S$, where $S$ is the surface tension.

Considering a small element of the thread subtending an angle $d\theta$ at the center, the outward force on this element is $dF = (2S)(R d\theta)$.

This outward force is balanced by the inward radial component of the tension $T$ in the thread. The inward force is $2T \sin\left(\dfrac{d\theta}{2}\right) \approx T d\theta$.

Equating the forces:

$T d\theta = 2S R d\theta$

$T = 2SR$

Substituting the given values $S = 0.030 \text{ N/m}$ and $R = 10^{-2} \text{ m}$:

$T = 2 \times 0.030 \times 10^{-2} = 6.0 \times 10^{-4} \text{ N}$