ABCDEF is a regular hexagon with each side 6 and each interior angle as 120

Triangle ABF is an isosceles triangle with AB = AF = 6, ∠ABF = ∠AFB = 30, ∠BAF = 120

If we drop a perpendicular from A to BF, ∠BAA' = ∠FAA' = 60 and BA' = A'F
Thus, triangle AA'B and AA'F are 30-60-90 triangles
∠i.e. ABA' = AFA' = 30, ∠BAA' = FAA' = 60

We know in 30-60-90 triangle, opposite to 90 is side 2x
i.e. \(2x = 6\)
\(x = 3\)

So, A'B = A'F = \(3\sqrt{3}\) and BF = CE = \(6\sqrt{3}\)

Perimeter of rectangle BCEF = \(2(6 + 6\sqrt{3}) = 12(1 + \sqrt{3}) = 12(2.732) = 32.784\)

Hence, option B