The product of the roots of the equation $$9x^2 - 18|x| + 5 = 0$$ is:

We start with the given quadratic expression that involves the modulus function:

$$9x^2-18|x|+5=0.$$

The absolute value $$|x|$$ creates two separate algebraic cases, because by definition

$$ |x|= \begin{cases} x & \text{when } x\ge 0,\\[4pt] -x & \text{when } x<0. \end{cases} $$

We therefore analyse each case separately and afterwards retain only those roots that satisfy the sign condition assumed while forming that case.

Case 1 : We assume $$x\ge 0,$$ so $$|x|=x.$$ Substituting this in the original equation gives

$$9x^2-18x+5=0.$$

To solve any quadratic of the form $$ax^2+bx+c=0,$$ we use the quadratic-formula

$$x=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}.$$

Here, $$a=9,\; b=-18,\; c=5.$$ The discriminant is computed first:

$$\Delta=b^2-4ac=(-18)^2-4(9)(5)=324-180=144.$$

Its square-root is $$\sqrt{144}=12.$$ Substituting in the formula:

$$ x=\dfrac{-(-18)\pm 12}{2\cdot 9} =\dfrac{18\pm 12}{18}. $$

This gives two roots

$$x_1=\dfrac{18+12}{18}=\dfrac{30}{18}=\dfrac{5}{3},\qquad x_2=\dfrac{18-12}{18}=\dfrac{6}{18}=\dfrac{1}{3}.$$

Both numbers are non-negative, so both belong to the present case and are valid.

Case 2 : We assume $$x<0,$$ so $$|x|=-x.$$ Substituting this into the original equation yields

$$9x^2-18(-x)+5=9x^2+18x+5=0.$$

Again we identify $$a=9,\; b=18,\; c=5$$ and apply the quadratic formula. First the discriminant:

$$\Delta=b^2-4ac=18^2-4(9)(5)=324-180=144,$$

and again $$\sqrt{\Delta}=12.$$ Thus

$$ x=\dfrac{-18\pm 12}{2\cdot 9} =\dfrac{-18\pm 12}{18}. $$

This furnishes

$$x_3=\dfrac{-18+12}{18}=\dfrac{-6}{18}=-\dfrac{1}{3},\qquad x_4=\dfrac{-18-12}{18}=\dfrac{-30}{18}=-\dfrac{5}{3}.$$

Both are negative, matching the assumption $$x<0,$$ hence both are admissible.

Collecting the four valid roots, we have

$$x_1=\dfrac{5}{3},\; x_2=\dfrac{1}{3},\; x_3=-\dfrac{1}{3},\; x_4=-\dfrac{5}{3}.$$

The product of all roots is

$$ x_1x_2x_3x_4 =\left(\dfrac{5}{3}\right)\!\left(\dfrac{1}{3}\right)\! \left(-\dfrac{1}{3}\right)\!\left(-\dfrac{5}{3}\right). $$

First multiply the two negative factors:

$$\left(-\dfrac{1}{3}\right)\!\left(-\dfrac{5}{3}\right)=\dfrac{5}{9}.$$

Next multiply the two positive factors:

$$\left(\dfrac{5}{3}\right)\!\left(\dfrac{1}{3}\right)=\dfrac{5}{9}.$$

Finally multiply these intermediate results:

$$\dfrac{5}{9}\times\dfrac{5}{9}=\dfrac{25}{81}.$$

Hence, the correct answer is Option B.