1. Set up the situation

The mass $m$ is attached to a spring of natural length $l_0$ and rotates on a frictionless horizontal table about the fixed end with angular speed $\omega$.

If the spring elongation is $x$, then the instantaneous length of the spring is $r = l_0 + x.$

2. Forces on the mass

The only horizontal force on the mass is the spring force, directed toward the fixed end: $F_{\text{spring}} = kx.$

Since the mass is moving in a circle of radius $r$, the required centripetal force is $F_{\text{centripetal}} = m\omega^2 r = m\omega^2 (l_0 + x).$

Thus, $kx = m\omega^2(l_0 + x).$

3. Solve for elongation $x$

Expand the right-hand side: $kx = m\omega^2 l_0 + m\omega^2 x.$

Bring the $x$-terms together: $kx - m\omega^2 x = m\omega^2 l_0.$

Factor out $x$: $x(k - m\omega^2) = m\omega^2 l_0.$

Hence, $x = \frac{m\omega^2 l_0}{k - m\omega^2}.$

4. Match with the options

This matches Option C: $\boxed{\frac{m\omega^2 l_0}{k - m\omega^2}}.$

5. Verification with stored answer

Stored correct answer: C
Derived answer: C
So they agree.