= Solution
The <Abelian field-strength chiral projection> is linear in the <vector superfield>, so isolate the terms containing the <gaugino> $\lambda$ and the $D$ <auxiliary field>. Replacing $x$ by $y-i\theta\sigma\bar\theta$ leaves these terms unchanged: the shift of the <gaugino> term would contain three barred <Grassmann variables>, and the shift of the $D$ term would contain three of each chirality, hence both vanish. Their contribution is therefore
$$
V_{\lambda,D}(y,\theta,\bar\theta)=(\bar\theta\bar\theta)\theta^\beta\lambda_\beta(y)
+\frac12(\theta\theta)(\bar\theta\bar\theta)D(y).
$$
The $2i(\sigma^\mu\bar\theta)_\alpha\partial_\mu$ part of the <supersymmetric covariant derivative> also adds a third barred factor, so it vanishes on these two terms. The ordinary <left Grassmann derivative>, with $\partial_\alpha(\theta\theta)=2\theta_\alpha$, gives
$$
D_\alpha V_{\lambda,D}=(\bar\theta\bar\theta)\{\lambda_\alpha(y)+\theta_\alpha D(y)\}.
$$
At fixed $y$, $\bar D=-\bar\partial$ and $\bar D^2(\bar\theta\bar\theta)=-4$. Thus the <chiral field-strength superfield> has
$$
\boxed{W_\alpha(y,\theta)=\lambda_\alpha(y)+\theta_\alpha D(y)+\text{terms containing }V_\mu\text{ or }\bar\lambda.}
$$
The remaining components come from the other independent terms in the <vector superfield>; linearity ensures they cannot alter the two coefficients just calculated. This proves the requested components without computing the omitted terms. The <gaugino> phase and the sign of the vector component are those in the printed <Wess-Zumino gauge> expansion.
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