Estimating the endpoint

Idea. The maximum can never exceed $\theta$, so it underestimates on average. Compute its mean and scale it up.

Step 1: distribution of the maximum. The maximum is at most $x$ exactly when all four observations are, so $$\mathbb{P}(\hat{\theta }\le x)={\left(\frac{x}{\theta }\right)}^4,0\le x\le \theta .$$

Step 2: its mean. Using the tail formula, $$\mathbb{E}[\hat{\theta }]={\int }_0^{\theta }\left(1-\frac{x^4}{{\theta }^4}\right)dx=\theta -\frac{\theta }{5}=\frac{4}{5}\theta .$$

Step 3. $\mathbb{E}[c\hat{\theta }]=\theta$ requires $c=\frac{5}{4}$.

Intuition. Four points split $[0,\theta ]$ into five gaps, and by symmetry each gap has mean $\frac{\theta }{5}$. The maximum misses the last gap, so it sits at $\frac{4}{5}\theta$ on average. With $n$ observations the correction is $\frac{n+1}{n}$.