Let $X = (X_1, X_2, X_3)$ be $\text{i.i.d.}$ $\text{Bin}(3,p)$ distributed and consider the estimator $\hat{p}(X) = \frac{X_1}{18} + \frac{X_2}{9} + \frac{X_3}{6}.$
First I had to show that $\hat{p}$ is unbiased, after some calculations I found: $E[\hat{p}] = p$
But the next question is that I have to show that this $\hat{p}$ is not sufficient and that I have to find a sufficient statistic for $p$.
Any help would be grateful. Thanks in advance.
multiplying the three densities you get
$$\binom{3}{x_1}\binom{3}{x_2}\binom{3}{x_3}p^{\sum x}(1-p)^{9-\sum x}=$$
$$=\underbrace{\binom{3}{x_1}\binom{3}{x_2}\binom{3}{x_3}}_{h(\mathbf{x})}\underbrace{(\frac{p}{1-p})^{\sum x}(1-p)^9}_{g[t(\mathbf{x}),p]}$$
Thus $T=\sum_i X_i$ is your sufficient estimator
In order to show that your $\hat{p}=\frac{x_1+2x_2+3x_3}{18}$ is not sufficient you can show that the conditional distribution $f_{X_1 X_2 X_3 | \hat{p}}$ is not independent from $p$ (a counterexample is enough)
COUNTEREXAMPLE
$$f(0,1,0|\hat{p}=\frac{2}{18})=...=1-p$$
that is not independent from $p$ and thus $\hat{p}=\frac{X_1+2X_2+3X_3}{18}$ is not sufficient for $p$
the denominator of the contitional probability is calculated considering that
$\hat{p}=\frac{2}{18}$ is achieved only in the following cases
$(0,1,0)$
$(2,0,0)$