I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $\rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $\rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$\LARGE\stackrel{\implies}{\vphantom{\prod_{n=1}^\Uparrow}}$
white circles (0 edges) $\to$ black diamonds (0 + 4 edges);
black circles (1 edge) $\to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $\to$ white $8$) and white pentagons go to black nonagons (white 5 $\to$ black $9$).
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.