I want to calculate the side lenght of $b$. I have two rectangles with one at 0° (screen) and I have one rectangle at 20° (turned image). With respect to the middle point. Both rectangles have a height of 6 and a width of 8.
Beceause the image is rotated, there will be black triangles in the corners of the screen. Now I want to calculate the lenght of $B$, as showed on the drawining. How can I do that?
Using the center where you've drawn the dot as the origin, and letting $c \approx .940$ and $s\approx .342$ denote the cosine and sine of $20$ degrees, respectively, the equation of the top horizontal line is $$ y = 3, $$ and the equation of the tilted top line is $$ \pmatrix{-s \\ c} \cdot \pmatrix{x\\y} = 3, $$ which is $$ -sx + cy - 3 = 0. $$ To compute the intersection, we plug in $y = 3$ into the second equation, getting $$ -sx + 3c - 3 = 0 \\ 3(c-1) = sx \\ \frac{3(c-1)}{s} = x $$ so that $$ (x, y) \approx (-0.530, 3) $$ is the intersection point on the top edge.
Doing the same for the left edge, whose equation is $$ x = -4, $$ we get $$ 4s + cy - 3 = 0 \\ cy = 3 - 4s \\ y = \frac{3 - 4s}{c} \approx \frac{3 - 4\cdot .342}{.940} \approx 1.737 $$ so that the left-hand intersection point is at $$ (x, y) \approx (-4, 1.737). $$ In short form:
Assuming inches, the top intersection is about .53 inches to the left of the middle of the top of the card; the left-hand intersection is about $1.74$ inches above the middle of the left edge of the card.
Now..about that length that you asked for: it's the distance between the two points. So we need to compute \begin{align} d &= \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \\ &= \sqrt{ (-0.53 - (-4))^2 + (3 - 1.737)^2 } \\ &\approx 3.693 \end{align}
If you want an exact formula, that'd be
\begin{align} d &= \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \\ &= \sqrt{ (\frac{3(c-1)}{s} - (-4))^2 + (3 - \frac{3 - 4s}{c})^2 } \\ &= \sqrt{ (\frac{3(c-1)}{s} + 4)^2 + (\frac{3c}{c} - \frac{3 - 4s}{c})^2 }\\ &= \sqrt{ (\frac{3(c-1)}{s} + \frac{4s}{s})^2 + (\frac{3c}{c} - \frac{3 - 4s}{c})^2 } \\ &= \sqrt{ (\frac{3(c-1) + 4s}{s})^2 + (\frac{3c - 3 + 4s}{c} )^2 } \\ &= \sqrt{ (\frac{3c- 3 + 4s}{s})^2 + (\frac{3c - 3 + 4s}{c} )^2 } \\ &= \sqrt{ \frac{(3c- 3 + 4s)^2}{s^2} + \frac{(3c - 3 + 4s)^2}{c^2} } \\ &= \sqrt{ \frac{(3c- 3 + 4s)^2 c^2}{s^2 c^2} + \frac{(3c - 3 + 4s)^2s^2}{s^2c^2} } \\ &= \sqrt{ \frac{(3c- 3 + 4s)^2 (c^2+s^2)}{s^2 c^2}} \\ &= \sqrt{ \frac{(3c- 3 + 4s)^2}{s^2 c^2}} \\ &= \frac{|3c- 3 + 4s|}{s c} \end{align}
Plugging that into an online calculator, I get $d \approx 3.69378132923$.