= Solution
The expectation is
$$
\mathbb E_\theta[\widehat\theta_w]
=w_{AA}(1-\theta)^2+w_{BB}\theta^2
+2w_{AB}\theta(1-\theta).
$$
For this polynomial to equal $\theta$ for every $\theta\in(0,1)$, comparison of its constant, linear, and quadratic coefficients gives
$$
w_{AA}=0,
\qquad 2w_{AB}=1,
\qquad w_{BB}-2w_{AB}=0.
$$
Thus the unique choice is
$$
\boxed{\ w^*=(0,1,1/2),
\qquad
\widehat\theta_{w^*}=\frac{2n_{BB}+n_{AB}}{2n}.\ }
$$
For fly $i$, let $Z_i$ be half its number of $B$ alleles. Then $Z_i\in\{0,1/2,1\}$, the $Z_i$ are independent and identically distributed,
$$
\mathbb EZ_i=\theta,
\qquad
\operatorname{Var}(Z_i)=\frac{\theta(1-\theta)}2,
$$
and $\widehat\theta_{w^*}=n^{-1}\sum_iZ_i$. The <weak law of large numbers> gives $\widehat\theta_{w^*}\to\theta$ in probability. Hence \b[the estimator is unbiased and consistent].
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