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
$$
\int_5^\infty \theta f(\theta)\;d\theta = 6.
$$