In the sequel $[XYZ]$ will represent the area of a triangle $XYZ$. Denote by $S$ the area of the polygon $A_1A_2...A_{2010}$.
Notice that $[OA_1A_2] = [OA_2A_3] = \dots = [OA_{2010}A_1] = \frac{1}{2010}S$, since the polygon $A_1A_2...A_{2010}$ is given as being regular. Then for each $k = 1, 2, ..., 2009$ we have
$$\frac{[OB_k B_{k+1}]}{[OA_k A_{k+1}]} = \frac{OB_k \cdot OB_{k+1}}{OA_k \cdot OA_{k+1}} = \frac{1}{k(k+1)},$$
as well as
$$\frac{[OB_{2010}B_1]}{[OA_{2010}A_1]} = \frac{OB_{2010} \cdot OB_1}{OA_{2010} \cdot OA_1} = \frac{1}{2010},$$
since the ratio of the areas of two triangles sharing a same angle is equal to the ratio of the products of the corresponding sides.
Therefore denoting by $T$ the area of the polygon $B_1B_2...B_{2010}$, we have
$$\begin{align*} T &= \sum_{k=1}^{2009} [OB_k B_{k+1}] + [OB_{2010} B_1] \\ &= \frac{S}{2010} \left( \sum_{k=1}^{2009} \frac{1}{k(k+1)} + \frac{1}{2010} \right) \\ &= \frac{S}{2010} \left( \sum_{k=1}^{2009} \left( \frac{1}{k} - \frac{1}{k+1} \right) + \frac{1}{2010} \right) \\ &= \frac{S}{2010} \left( \left( 1 - \frac{1}{2010} \right) + \frac{1}{2010} \right) = \frac{S}{2010}, \end{align*}$$
hence $\frac{T}{S} = \frac{1}{2010}$.