= Solution
Use the <supertranslation-invariant superspace coframe>
$$
\boxed{\Pi^m=dX^m+i\bar\theta\gamma^m d\theta,
\qquad\Psi^\alpha=d\theta^\alpha.}
$$
It is a basis because the change from $(dX,d\theta)$ is triangular with identity diagonal blocks. Under a rigid <supertranslation>, $\delta\theta=\epsilon$, $\delta X^m=-i\bar\epsilon\gamma^m\theta$ and $d\epsilon=0$. Therefore $\delta\Pi^m=-i\bar\epsilon\gamma^m d\theta+i\bar\epsilon\gamma^m d\theta=0$, and $\delta d\theta=0$.
Writing wedge products explicitly, the three-form is
$$
H_3=\Pi^m\wedge d\bar\theta\gamma_m d\theta.
$$
Both factors are <supertranslation> invariant. Under a <Lorentz transformation>, $\Pi^m$ and the bilinear in $d\theta$ transform as vectors, so their contraction is invariant. Thus $H_3$ is invariant under the <Super-Poincaré group>.
The <Maurer-Cartan equation> is $d\Pi^m=i\,d\bar\theta\gamma^m d\theta$. Since the other factor is closed,
$$
dH_3=i(d\bar\theta\gamma^m d\theta)
\wedge(d\bar\theta\gamma_m d\theta).
$$
The required four-dimensional <Fierz rearrangement> identity, which may be stated without proof, is
$$
(C\gamma^m)_{(\alpha\beta}(C\gamma_m)_{\gamma\delta)}=0.
$$
Differentials of odd <Grassmann variables> commute in the graded exterior algebra. Thus the coefficient of the product of four $d\theta$ factors is the fully symmetric spinor-index part appearing in this identity. It vanishes, proving
$$
\boxed{dH_3=0.}
$$
This is the <closed three-form on four-dimensional superspace>; it should not be confused with the different closed four-form used for a membrane coupling in Question 10.
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