The prior density is $p(\theta)=\theta e^{-\theta}$ for $\theta>0$, and the likelihood function is $ L(\theta) = \begin{cases} \frac 1 \theta & \text{for }\theta\ge5, \\ \\ \\ 0 & \text{otherwise}. \end{cases} $ Multiplying them, you get $ g(\theta)=p(\theta) L(\theta) = \begin{cases} e^{-\theta} & \text{for }\theta\ge5, \\ \\ \\ 0 & \text{otherwise}. \end{cases} $ Then we have $ \int_5^\infty g(\theta)\;d\theta = e^{-5},\text{ so }\int_5^\infty e^5 g(\theta)\;d\theta = 1. $ Therefore $f(\theta)= \begin{cases} e^{5-\theta} & \text{if }\theta>5 \\ \\ \\ 0 & \text{if }\theta<5 \end{cases}$ is the posterior probability density function. With squared-error loss, the Bayes estimate is just the posterior expected value $ \int_5^\infty \theta f(\theta)\;d\theta = 6. $