A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point (0, -α) to the parabola x² = -4ay, where a > 0. Let L be the line passing through (0, -α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r : s = 1 : 16, then the value of 24a is __________.

Given the parabola is:
$x^2=-4ay$
where $a>0$.

The length of the latus rectum $r$ of this parabola is:
$r=4a$

The directrix of the parabola $x^2=-4ay$ is the horizontal line:
$y=a$

The line $L$ passes through the point $(0,-\alpha )$ and is parallel to the directrix ($y=a$). Thus, $L$ is the horizontal line:
$y=-\alpha$

Line $L$ intersects the parabola at two points $A$ and $B$. Substituting $y=-\alpha$ into the parabola's equation gives:
$x^2=-4a(-\alpha )=4a\alpha$
So, the x-coordinates of $A$ and $B$ are:
$x=\pm 2\sqrt{a\alpha }$

The length of the line segment $AB$ is:
$AB=4\sqrt{a\alpha }$
Thus, the square of the length of $AB$, denoted by $s$, is:
$s=A{B}^2=16a\alpha$

We are given the ratio:
$r:s=1:16$
Substituting the expressions for $r$ and $s$:
$\frac{4a}{16a\alpha }=\frac{1}{16}$
Simplifying this relation:
$\frac{1}{4\alpha }=\frac{1}{16}\Rightarrow \alpha =4$
Thus, the point from which the normal is drawn is $(0,-4)$.

Now, let's write the equation of a normal to the parabola $x^2=-4ay$ in terms of its slope $m$.
For a parabola of the form $x^2=4Ay$, the equation of the normal in slope form is:
$y=mx-2A-A{m}^2$
Substituting $A=-a$, the equation of the normal with slope $m$ to $x^2=-4ay$ is:
$y=mx+2a+a{m}^2$

We are given that the normal has slope:
$m=\frac{1}{\sqrt{6}}$
Substituting this slope $m$ into the normal equation:
$y=\frac{1}{\sqrt{6}}x+2a+a\left(\frac{1}{6}\right)=\frac{1}{\sqrt{6}}x+\frac{13a}{6}$

Since this normal passes through the point $(0,-\alpha )=(0,-4)$, we substitute $x=0$ and $y=-4$:
$-4=\frac{1}{\sqrt{6}}\left(0\right)+\frac{13a}{6}$
$\Rightarrow \frac{13a}{6}=-4$

To get the value of $24a$:
Since $a>0$, and we require the magnitude/absolute values or correct standard geometry:
The distance from vertex to directrix or coordinates leads to the relation where $\alpha >0$ corresponds to the line below the x-axis, and normal parameters yield:
$24a=4\times \left(6a\right)$
Taking the positive coordinate/standard geometry context where $y=-mx-2a-a{m}^3$ parameterization is handled, we get:
$2a+a{m}^2=\alpha$
Substituting $\alpha =4$ and $m=\frac{1}{\sqrt{6}}$:
$2a+a\left(\frac{1}{6}\right)=4$
$\Rightarrow \frac{13a}{6}=4$
But using the standard parametric normal equation for $x^2=-4ay$ where the normal at $(2at,-a{t}^2)$ is:
$y+a{t}^2=\frac{1}{t}(x-2at)$
$\Rightarrow y=\frac{1}{t}x-2a-a{t}^2$
Here the slope of the normal is $m=\frac{1}{t}=\frac{1}{\sqrt{6}}\Rightarrow t=\sqrt{6}$.
Substituting $t=\sqrt{6}$ into the equation of the normal:
$y=\frac{1}{\sqrt{6}}x-2a-6a$
$\Rightarrow y=\frac{1}{\sqrt{6}}x-8a$

This normal passes through $(0,-\alpha )=(0,-4)$:
$-4=-8a$
$\Rightarrow 8a=4$
Multiply both sides by 3 to find $24a$:
$24a=12$

Thus, the value of $24a$ is 12.