The problem can be represented diagrammatically as follows:

Here AB is the longest wire (30m) and (OC) is the shortest at 6m. LN is the supporting wire, 18m from the middle.

We can see that BC = half of AB = 50m.

The shape of the parabola opens upwards, therefore the equation is $x^2 = 4ay$.

We can see that coordinates of A is $(50, 24)$ (i.e, $y$ coordinate is $AB - OC = 30 - 6 = 24$)

Since $A$ is a point on the parabola, $50^2 = 4\times a \times 24^2 \rightarrow a = \large\frac{625}{24}$

Therefore, the equation of the parabola is $x^2 = 4\times\large\frac{625}{24}$$y \rightarrow 6x^2 = 625y$

$x$ coordinate at point L $=18 \rightarrow 6 \times 18^2 = 625 y \rightarrow y $ coordinate at point L $ = \large\frac{6\times 18^2}{625}$$ \approx 3.11$

We need to calculate LN, which is $3.11 + 6 = 9.11m$.