= Solution
{c}
For an observed <random sample> with all $x_i>0$, which happens <almost surely>, <independence> gives the <likelihood function>
$$
L(\theta)=\prod_{i=1}^n\frac{2x_i}{\theta}e^{-x_i^2/\theta}=2^n\left(\prod_{i=1}^nx_i\right)\theta^{-n}\exp\left(-\frac{S}{\theta}\right),\qquad S=\sum_{i=1}^nx_i^2>0.
$$
Ignoring a term independent of $\theta$, the <log-likelihood> and its <derivative> are
$$
\ell(\theta)=-n\log\theta-\frac S\theta+C,\qquad \ell'(\theta)=\frac{S-n\theta}{\theta^2}.
$$
This <derivative> is positive for $0<\theta<S/n$ and negative for $\theta>S/n$. Thus there is a unique <global maximum>, and the <maximum-likelihood estimator> is
$$
\boxed{\widehat\theta(\mathbf X)=\frac1n\sum_{i=1}^nX_i^2.}
$$
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