The figure below is formed by a square ABCD, equilateral triangle ABE and triangles BDG and BEG. AFE and BFD are straight lines.

1. Identify the angles of the square and the equilateral triangle at vertex B

Since $ABCD$ is a square, $\angle ABC = 90^\circ$.
Since $\triangle ABE$ is an equilateral triangle, $\angle ABE = 60^\circ$.

2. Calculate $\angle CBE$

From the figure's description, $E$ lies inside the square (as $AE$ intersects the diagonal $BD$).
$\angle CBE = \angle ABC - \angle ABE$
$\angle CBE = 90^\circ - 60^\circ = 30^\circ$