There is no factor of 2. The Dirac-Bergmann analysis$^{\dagger}$ goes as follows. The second-class constraints are $$ \chi^i ~=~\pi^i - \frac{1}{2}\epsilon^{ij} A_j, \qquad i~\in~\{1,2\}. $$ The matrix of Poisson brackets of second-class constraints is $$ \Delta^{ij}(x,y)~:=~\{\chi^i(x),\chi^j(y)\}~=~-\epsilon^{ij}\delta^2(x-y), $$ so the inverse matrix is $$ (\Delta^{-1})_{ij}(x,y)~=~-(\epsilon^{-1})_{ij}\delta^2(x-y). $$ The Dirac bracket becomes $$ \{A_i(x),A_j(y)\}_D~=~(\epsilon^{-1})_{ij}\delta^2(x-y). $$

References:

1. G.V. Dunne, arXiv:hep-th/9902115.

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$^{\dagger}$ Ref. 1 implicitly mentions between eqs. (70)-(71) a shortcut via the Faddeev-Jackiw method.

There is no factor of 2. The Dirac-Bergmann analysis$^1$ goes as follows. The second-class constraints are $$ \chi^i ~=~\pi^i - \frac{1}{2}\epsilon^{ij} A_j, \qquad i~\in~\{1,2\}. $$ The matrix of Poisson brackets$^2$ of second-class constraints is $$ \Delta^{ij}(x,y)~:=~\{\chi^i(x),\chi^j(y)\}~=~-\epsilon^{ij}\delta^2(x-y), $$ so the inverse matrix is $$ (\Delta^{-1})_{ij}(x,y)~=~-(\epsilon^{-1})_{ij}\delta^2(x-y). $$ The Dirac bracket becomes $$ \{A_i(x),A_j(y)\}_D~=~-(\epsilon^{-1})_{ij}\delta^2(x-y). $$

References:

1. G.V. Dunne, arXiv:hep-th/9902115.

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$^1$ Ref. 1 implicitly mentions between eqs. (70)-(71) a shortcut via the Faddeev-Jackiw method.

$^2$ To go from brackets to commutators, multiply with $i\hbar$.