Work done = Change in potential energy = U(final) - U(initial) , and,

U = k{ (Q1.Q2)/r1 +(Q2.Q3)/r2 +(Q3.Q1)/r3 }, where k = 9x10^9

**Initially**, when charges are kept in equilateral triangle of side 1 m :

Since triangle is equilateral, r1 = r2 = r3 = r = 1 m

Also, Q1 = 1C , Q2 = 2C, Q3 = 3C

Therefore,

U(initial) = k{1x2 + 2x3 + 3x1}/1 = 11k

Now

**For position A :**

All three charges are in a new equilateral triangle of side r = 0.5 m

Therefore,

U(final) for A = k{1x2 + 2x3 + 3x1}/0.5 = 22k

Hence, Work done in A = U(final) - U(initial) = 22k - 11k = 11k

**For position B :**

All three charges are in a new equilateral triangle of side r = 0.5 m

U(final) for B = k{1x2 + 2x3 + 3x1}/0.5 = 22k

hence, Work done in B = U(final) - U(initial) = 22k - 11k = 11k

From above, clearly, Work in A = Work in B

Hence, **option C is true.**

Above was the complete approach, but since the options do not deal with exact data, we can use another approach which will be much shorter for this question.

**As we know, Work done of a conservative force is path independent, i.e, it only depends on initial and final conditions of the system**.

*In our question Coulomb's Force is mentioned, which is a conservative force, hence we will just check for initial and final conditions.*

Initially, system is an equilateral triangle of side 1 m.
Finally, system is an equilateral triangle of side 0.5 m for both A and B.
Hence, as these conditions are same, the work done will also be the same, irrespective of the path followed to achieve the final condition in method A or method B.
thus,

**Work in A = Work in B**