Five rays $\overrightarrow{OA}$,$\overrightarrow{OB}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$, and $\overrightarrow{OE}$  radiate in a clockwise order from $O$ forming four non-overlapping angles such that $\angle EOD = 2\angle COB$, $\angle COB = 2\angle BOA$, while $\angle DOC = 3\angle BOA$. If $E$, $O$, $A$ are collinear with $O$ between $A$ and $E$, what is the degree measure of $\angle DOB?$
1. Let $\angle AOB = x^\circ$. According to the problem, we have the following relationships:
   - $\angle COB = 2x^\circ$
   - $\angle DOC = 3x^\circ$
   - $\angle EOD = 2 \cdot \angle COB = 2 \cdot 2x^\circ = 4x^\circ$

2. Since $E$, $O$, and $A$ are collinear with $O$ between $A$ and $E$, the sum of the angles around point $O$ must be $180^\circ$. Therefore, we can write the equation:
   \[
   \angle AOB + \angle COB + \angle DOC + \angle EOD = 180^\circ
   \]
   Substituting the given angle measures, we get:
   \[
   x^\circ + 2x^\circ + 3x^\circ + 4x^\circ = 180^\circ
   \]

3. Simplify the equation:
   \[
   10x^\circ = 180^\circ
   \]

4. Solve for $x$:
   \[
   x = \frac{180^\circ}{10} = 18^\circ
   \]

5. Now, we need to find the measure of $\angle DOB$. According to the problem, $\angle DOB$ is the sum of $\angle DOC$ and $\angle COB$:
   \[
   \angle DOB = \angle DOC + \angle COB
   \]

6. Substitute the values of $\angle DOC$ and $\angle COB$:
   \[
   \angle DOC = 3x^\circ = 3 \cdot 18^\circ = 54^\circ
   \]
   \[
   \angle COB = 2x^\circ = 2 \cdot 18^\circ = 36^\circ
   \]

7. Therefore:
   \[
   \angle DOB = 54^\circ + 36^\circ = 90^\circ
   \]

The final answer is $\boxed{90^\circ}$