Please help.i want solution to a problem regarding surface tension?

The work done is against the external atmospheric pressure p0 and the surface tension.

The latter is 4γ/r ( 2γ/r for each soap-air interface)

Then

W = ∫ p dV = ∫ (p0 + 4 γ/r) 4πr^2 dr,

integrated between r=0 and r=R.

Thus

W = 4/3 πR^3 p0 + 8πγR^2

Just plug in the numbers given ( convert cm to m first).

You can also calculate this expression for R = 0.20 m. Subtracting the work for R= 0.10 m will give the additional work.

Note that this is the total work done. Maybe the exercise wants you to just calculate the work against surface tension. In that case only the second term is needed of course.

Work done in forming a soap bubble is 8pieR^2S (S is surface tension and R is radius).Just substitute and find out.

When the radius is doubled R becomes 2R so the eqtn becomes,

8pie(2R)^2S

Also bfore calculating dont forget to convert 10cm to metres i.e 10x100m bcoz both the values must be in the same units,

Hope this helps...:)