Decide if a language has a word of a given size
There are two cases:


$L$ is empty. In this case, $L' = \emptyset$ is trivially decidable.
$L$ is non-empty. Let $m$ be the minimum length of a word in $L$. Then $L'$ consists of all words of length at least $m$, and is again trivially decidable (in constant time!).


As you can see, you never actually need an algorithm for $L$.



Similarly, the following language is always decidable:

$$L'' = \{w \in \Sigma^* \mid \text{ there exists a word $w' \in L$ such that $|w'| \geq |w|$}\}.$$

There are now three cases:


$L$ is empty. In this case, $L'' = \emptyset$ is trivially decidable.
$L$ is infinite. In this case, $L'' = \Sigma^*$ is again triviall decidable.
$L$ is finite. Let $M$ be the maximum length of a word in $L$. Then $L''$ consists of all words of length at most $M$, and is again trivially decidable (in constant time).




These are examples of non-constructive proofs, which you might not like. Instead of starting a discussion here, I refer you to this question.