= Solution
Use <left Grassmann derivatives> and the usual contractions $\partial^2=\partial^\alpha\partial_\alpha$ and $\bar\partial^2=\bar\partial_{\dot\alpha}\bar\partial^{\dot\alpha}$. Set $a=\theta^1,b=\theta^2,c=\bar\theta^{\dot1},d=\bar\theta^{\dot2}$. The printed epsilon convention gives
$$
\theta_1=-b,\quad\theta_2=a,\qquad
\bar\theta_{\dot1}=-d,\quad\bar\theta_{\dot2}=c,
\qquad \theta\theta=-2ab,\quad\bar\theta\bar\theta=2cd.
$$
The reversal of the barred contraction is essential. The <left Grassmann derivative> obeys the <graded Leibniz rule>, so $\partial_a(ab)=b$ but $\partial_b(ab)=-a$. With the index-raising convention,
$$
\partial^2=2\partial_b\partial_a,\qquad
\bar\partial^2=2\partial_c\partial_d.
$$
Applying these to the displayed <Grassmann algebra> monomials gives \b[$A=B=-4$]. Since every antisymmetric two-index product is proportional to epsilon, the $12$ and $\dot1\dot2$ components give
$$
\theta^\alpha\theta^\beta=-\frac12\epsilon^{\alpha\beta}(\theta\theta),\qquad
\bar\theta^{\dot\alpha}\bar\theta^{\dot\beta}=\frac12\epsilon^{\dot\alpha\dot\beta}(\bar\theta\bar\theta).
$$
Thus \b[$C=-2$ and $D=2$].
In the remaining contraction, moving the first barred <Grassmann variable> past the second unbarred one introduces a minus sign. Inserting the two spinor identities then gives
$$
\begin{aligned}
(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta)
&=\frac14(\theta\theta)(\bar\theta\bar\theta)\epsilon^{\alpha\beta}\epsilon^{\dot\alpha\dot\beta}
\sigma^\mu_{\alpha\dot\alpha}\sigma^\nu_{\beta\dot\beta}\\
&=\frac14(\theta\theta)(\bar\theta\bar\theta)\operatorname{Tr}(\sigma^\mu\bar\sigma^\nu)\\
&=\frac12(\theta\theta)(\bar\theta\bar\theta)\eta^{\mu\nu}.
\end{aligned}
$$
The trace normalization is the one explicitly supplied in the paper. These <two-component superspace contraction signs> therefore give
$$
\boxed{(A,B,C,D,E)=(-4,-4,-2,2,2).}
$$
As a direct check in the mostly-minus convention, $\sigma^0=I$ gives $(ac+bd)^2=-2abcd$, whereas $(\theta\theta)(\bar\theta\bar\theta)=-4abcd$. Their ratio is $+1/2$. A convention with $\operatorname{Tr}(\sigma^\mu\bar\sigma^\nu)=-2\eta^{\mu\nu}$ would change this last metric-relative sign; it is not the printed trace convention.
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