If $L$ is a regular language then so is $\sqrt{L}=\{w:ww\in L\}$
Here is how to implement your solution. Let $A = \langle Q, q_0, F, \delta \rangle$ be a DFA for $L$. We will construct an NFA $A' = \langle Q', q'_0, F', \delta' \rangle$ as follows:


$Q' = \{q'_0\} \cup Q^3$. The state $(q_1,q_2,q_3)$ means that we have guessed that when $A$ finishes reading the first copy of $w$, it will be in state $q_1$; the first copy of $A$, started at $q_0$, is at state $q_2$; and the second copy of $A$, started at $q_1$, is at state $q_3$.
$F' = \{(q_1,q_1,q_2) : q_1 \in Q, q_2 \in F\}$. Thus we accept if the first copy of $A$ is in the guessed state, and the second copy of $A$ is at an accepting state.
$\delta'(q'_0,\epsilon) = \{(q,q_0,q) : q \in Q\}$. This initializes the simulation of the two copies of $A$.
$\delta'((q_1,q_2,q_3),a) = \{(q_1,\delta(q_2,a),\delta(q_3,a))\}$. This simulates both copies of $A$, while keeping the guessed state.


We leave the reader the formal proof that $L(A') = \sqrt{L(A)}$.



Here is another solution, which creates a DFA. We now run $|Q|$ copies of $A$ in parallel, starting at each state of $A$:


$Q' = Q^Q$.
$q'_0 = q \mapsto q$, the identity function.
$\delta'(f,a) = q \mapsto \delta(q,a)$. 
$F' = \{ f \in Q' : f(f(q_0)) \in F \}$.


What is the meaning of the condition $f(f(q_0)) \in F$? After reading a word $w$, the automaton $A'$ is in a state $f$ given by $f(q) = \delta(q,w)$. Thus $f(f(q_0)) = \delta(\delta(q_0,w),w) = \delta(q_0,w^2)$.