1. Work done in expanding a soap bubble

For a soap bubble, there are two surfaces, so the surface energy is

$U = 2T \cdot 4\pi r^2 = 8\pi T r^2$

Hence, the work done in increasing radius from $r_1$ to $r_2$ is

$W = 8\pi T (r_2^2 - r_1^2)$

2. Given data

$T = 3.5 \times 10^{-2}\, \text{N m}^{-1}$ $r_1 = 10\, \text{cm} = 0.1\, \text{m}$ $r_2 = 20\, \text{cm} = 0.2\, \text{m}$

Now,

$r_2^2 - r_1^2 = (0.2)^2 - (0.1)^2 = 0.04 - 0.01 = 0.03$

3. Substitute into formula

$W = 8 \times \frac{22}{7} \times 3.5 \times 10^{-2} \times 0.03$

Since

$3.5 = \frac{7}{2}$

we get

$8 \times \frac{22}{7} \times \frac{7}{2} \times 10^{-2} \times 0.03$

Cancel $7$:

$= 8 \times 11 \times 10^{-2} \times 0.03$

$= 88 \times 3 \times 10^{-4}$

$= 264 \times 10^{-4}\, \text{J}$

4. Final answer

The required value in the blank is

$\boxed{264}$

So, the work done is

$\boxed{264 \times 10^{-4}\, \text{J}}$

5. Comparison with stored answer

Stored correct answer = $264$

Our derived answer also = $264$.

Hence, the answer agrees with the stored answer.