We are provided the language :
$$L=\{w| |w| \text{ is prime} \}^*$$
Let us investigate the type of strings in $L$. We see that $L$ has such strings whose length is either zero or can be expressed as a sum of prime numbers. i.e. if $x \in L$ then we have :
$$|x| = \begin{cases}
0 \text{ or}\\
\Sigma p_i \text{ where $p_i$ $\in$ Set of all prime numbers}\\
\end{cases}$$
Now let us consider first the set of all strings of even length. i.e. $|x| = 2k , \text{$k \in \mathbb{N}$ and $k\geq 0$}$
So given any string of even length, we can write it as a concatenation of $0$ or more number strings of length $2$, but we see that $2$ is a prime number. So any string of even length $\in L$ and can be expressed by the regular expression :
$$((a+b)(a+b))^*$$
From the above expression, we can generate strings of length $0,2,4,...$ $\tag 1$
Let us consider the strings of odd length.
Let us consider the string of length $1$. But $1$ is not a prime number. So strings of length $1$ cannot be present in $L$. So we are left with the odd numbers $3,5,7,...$ or
$$|x|= 2\lambda+1, \text{$\lambda \in \mathbb{N}$ and $\lambda\geq 1$} \tag 2$$
$$\implies |x| = 3+ (2\lambda+1 -3) = 3+ (2\lambda -2)=3+ 2(\lambda -1)$$
$$|x| = 3+2\mu \text{ , $\mu \in \mathbb{N}$ and $\mu\geq 0$}$$
Now $3$ and $2$ are prime numbers. So any odd length string of length greater than $1$ can be represented by the concatenation of a string of prime length $3$ followed by zero or more numbers of strings of prime length $2$, which can be represented by the regular expression:
$$(a+b)(a+b)(a+b)((a+b)(a+b))^*$$
From the above two calculations we can say that L:
$$L = ((a+b)(a+b))^*+(a+b)(a+b)(a+b)((a+b)(a+b))^*$$
This is how we suggest that $L$ is a regular language.
Now as far as the language is concerned, we see from $(1)$ and $(2)$ we see that strings in $L$ have lengths $0,2,3,4,5,6,...$, which is simply :
$$L= \epsilon + (a+b)(a+b)(a+b)^*$$