= Solution
Write $A=(X^TX)^{-1}X^T$. Since $AX=I_p$,
$$
\widehat\theta=\theta+A\epsilon.
$$
The errors have mean zero and <covariance matrix> $\sigma^2I_n$. Therefore
$$
\boxed{\mathbb E\widehat\theta=\theta,\qquad
\operatorname{Cov}(\widehat\theta)=\sigma^2AA^T
=\sigma^2(X^TX)^{-1}.}
$$
Thus $\widehat\theta$ is an <unbiased estimator>. In addition, the <multivariate normal distribution> is preserved by <linear map>[linear maps], so $\widehat\theta$ is <multivariate normal distribution>[multivariate normal] with the displayed mean and <covariance matrix>.
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