> I wonder if the flaw can be removed with this shape.

interesting!  i can show that no perfect hexagon is possible,
i assume that's what you're asking.
however, it is possible to make hexagonal annuli with arbitrarily
small inner hexagon and arbitrarily large outer hexagon.

now that i think about it some more, i'm less excited about it.

4 fins make a 60 degree parallelogram with edges 3 + sqrt(3) and
2 + 2 sqrt(3).  and these can be packed into a parallelogram
with edges  m(3 + sqrt(3))  and  n(2 + 2 sqrt(3)) , for any
positive integers  m  and  n .  note that 12 copies of an  a x b
60 degree parallelogram make a hexagonal annulus where the outer
hexagon has edge  a + b  and the inner hexagon has edge  |a - b| .
so the arbitrarily good packing comes from making the difference
m(3 + sqrt(3)) - n(2 + 2 sqrt(3))  small.

to show that a perfect hexagon is impossible, suppose its edge is  s .
note that  s = a + b sqrt(3)  for some integers  a  and  b .
then its area is  3/2 sqrt(3) s^2 .  the fin itself has area
((1 + sqrt(3))^2 + 1 (2 + sqrt(3)))/2 = (6 + 3 sqrt(3))/2.
now suppose that there are  n  of them in the perfect packing.
equating areas, we get  (6 + 3 sqrt(3)) n / 2  = 3/2 sqrt(3) s^2,
which gives  (2 + sqrt(3)) n = sqrt(3) s^2 .  multiply by  sqrt(3)
to get  (3 + 2 sqrt(3)) n = 3 s^2 .

but now take the conjugate of this equation.  if  s = a + b sqrt(3) ,
then it says  (3 - 2 sqrt(3)) n = 3 (a - b sqrt(3))^2 , which is
impossible, since the left hand side is negative!