find the shortest path from A to B Consider the L-shaped brick in the diagram. If an ant starts from $A$, find the minimum distance it has to travel along the surface to $B$. 

The answer is $\frac {3\sqrt{5}}2$ .
 A: As @ChristianBlatter pointed out in a comment, the Euclidean, 3-dimensional distance between $A$ and $B$ is larger than your given answer. Here is a view of your solid from the top, with the side containing point $B$ "folded" up.

As you can see, the two coordinate distances in the horizontal plane are $2$ and $3$, and the vertical distance is $1$. Therefore,
$$AB=\sqrt{2^2+3^2+1^2}=\sqrt{14}\approx 3.74166$$
which is larger than your given answer $\frac{3\sqrt 5}2\approx 3.3541$. So your given answer is clearly wrong.
For a better answer, let's consider two cases. In the first case, we travel from $A$ to $B$ first on the top L-shaped surface of the solid and we do not intersect the edge that is below and to the right of $A$ in your diagram. Then travelling on the solid's surface is mostly on the L-shaped tope and we get a similar diagram as the one above.

It is clear that the shortest path goes to the inside corner of the L, and the length of that shortest path is $\sqrt 2+\sqrt{10}\approx 4.57649$.
In the second case we do intersect the edge that is below and to the right of $A$ in your diagram, and we do some travel on the solid's sides. Then to get a diagram we again "fold" the three relevant sides of the solid with part of the L-shaped top.

The top L does not completely fit into this diagram, but I'm sure you get the idea. Again it is clear that the shortest path goes to the inside corner of the L, and the length of that shortest path is $\sqrt 2+\sqrt{10}\approx 4.57649$.
I do not see any other path that has any chance of being shorter, so the apparent answer is

$$\sqrt 2+\sqrt{10}$$

