$L_n = \{x \in \Sigma^{*} | \exists w, y, z \in \Sigma^*, x = ywz, w^r = w, |w| = n \}$ Informally x is palindrome of length n. Find regex for n = 1
According to the description $w$ is a palindrome of length $n$, and not the complete string $x$. A palindrome is a string that will be the same when reversed. For each length $n$ there are of course finitely many palindromes (as there are only finitely many strings of that length).

The language consists of all palindromes with given length, with arbitrary strings before and after. The palindromes can be listed,  and "hard-wired" in the expression.

$L_1 = (0+1)^*(0+1)(0+1)^*$, 
$L_2 = (0+1)^*(00+11)(0+1)^*$, 
$L_3 = (0+1)^*(000+010+101+111)(0+1)^*$, ...

The complement is added to see how many strings are not of the given form. Any nonempty string will contain a palindrome of length 1 of course. Hence $L_1^c= \{\varepsilon\}$.

Strings that do not contain $00$ or $11$ must have alternating $0$ and $1$'s,hance the $(01)^*$ etc.