Consider a n-sided polygon.

In order for a diagonal to form we need to connect 2 vertices out of the n vertices on the polygon.
We can do this is $^nC_2$ ways

But there are $n$ such possibilities out of these which are sides of the polygon, we just need to subtract those.

Hence number of diagonals in a n-sided polygon = $^nC_2 - n = \frac{n(n-1)}{2*1} - n = \frac{n(n-1) - 2n}{2} = \frac{n^2 - 3n}{2} = \frac{n(n-3)}{2}$

Hence number of diagonals in a n-sided polygon (Hexagon) = $\frac{n(n-3)}{2} = \frac{6(6-3)}{2} = \frac{6*3}{2} = 3*3 = \textbf{9}$

Hence, Answer is D