Let one root of the quadratic equation in $x$: $(k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

Let the roots of the given quadratic equation be $\alpha$ and $2\alpha$.

Sum of the roots:
$\alpha + 2\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27}$
$3\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27} \Rightarrow \alpha = \dfrac{-3(k-1)}{k^2 - 15k + 27}$

Product of the roots:
$\alpha \cdot 2\alpha = \dfrac{18}{k^2 - 15k + 27}$
$2\alpha^2 = \dfrac{18}{k^2 - 15k + 27} \Rightarrow \alpha^2 = \dfrac{9}{k^2 - 15k + 27}$

Substituting the value of $\alpha$ from the sum into the product equation:
$\left( \dfrac{-3(k-1)}{k^2 - 15k + 27} \right)^2 = \dfrac{9}{k^2 - 15k + 27}$

$\dfrac{9(k-1)^2}{(k^2 - 15k + 27)^2} = \dfrac{9}{k^2 - 15k + 27}$

Since $k^2 - 15k + 27 \neq 0$, we can simplify to:
$(k-1)^2 = k^2 - 15k + 27$

$k^2 - 2k + 1 = k^2 - 15k + 27$

$13k = 26 \Rightarrow k = 2$

The equation of the parabola is $y^2 = 6kx$. Substituting $k = 2$, we get:
$y^2 = 12x$

The length of the latus rectum of a parabola $y^2 = 4ax$ is $4a$, which is the coefficient of $x$.

Therefore, the length of the latus rectum is $12$.

Answer: $12$