Given that \(x\) varies directly as \(\sqrt{n}\), we can write this relationship as:

x = k \cdot \sqrt{n},\)

where \(k\) is the constant of variation.

We are given that when \(x = 9\) and \(n = 9\), the equation holds true:

9 = k \cdot \sqrt{9}.\)

Solving for \(k\):

k = \frac{9}{\sqrt{9}} = \frac{9}{3} = 3.\)

Now that we have the value of \(k\), we can use it to find \(x\) when \(n = \frac{17}{9}\):

x = 3 \cdot \sqrt{\frac{17}{9}}.\)

Simplifying under the square root:

x = 3 \cdot \sqrt{\frac{17}{9}} = 3 \cdot \frac{\sqrt{17}}{\sqrt{9}} = 3 \cdot \frac{\sqrt{17}}{3} = \sqrt{17}.\)

So, the correct answer is:

D. \, \sqrt{17}.\)