1. **Calculate the area of quadrilateral $ABCD$:**
- The area of $ABCD$ can be divided into the sum of the areas of triangles $ABM$, $BCM$, $CDM$, and $DAM$.
- The area of each triangle can be calculated using the formula for the area of a triangle:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \times \sin(\text{angle})$$
- Given $m(\widehat{AMB}) = 60^\circ$, we have:
$$[ABM] = \frac{1}{2} \times AM \times BM \times \sin(60^\circ) = \frac{1}{2} \times AM \times BM \times \frac{\sqrt{3}}{2} = \frac{AM \times BM \times \sqrt{3}}{4}$$
- Similarly, we can calculate the areas of the other triangles:
$$[BCM] = \frac{1}{2} \times BM \times CM \times \sin(60^\circ) = \frac{BM \times CM \times \sqrt{3}}{4}$$
$$[CDM] = \frac{1}{2} \times CM \times DM \times \sin(60^\circ) = \frac{CM \times DM \times \sqrt{3}}{4}$$
$$[DAM] = \frac{1}{2} \times DM \times AM \times \sin(60^\circ) = \frac{DM \times AM \times \sqrt{3}}{4}$$
- Summing these areas, we get:
$$[ABCD] = [ABM] + [BCM] + [CDM] + [DAM] = \frac{AM \times BM \times \sqrt{3}}{4} + \frac{BM \times CM \times \sqrt{3}}{4} + \frac{CM \times DM \times \sqrt{3}}{4} + \frac{DM \times AM \times \sqrt{3}}{4}$$
- Since $M$ is the intersection of the diagonals, we can express the area in terms of the diagonals $AC$ and $BD$:
$$[ABCD] = \frac{AC \times BD \times \sqrt{3}}{4}$$

2. **Calculate the area of quadrilateral $O_1O_2O_3O_4$:**
- The points $O_1, O_2, O_3, O_4$ are the circumcenters of triangles $ABM$, $BCM$, $CDM$, and $DAM$, respectively.
- The quadrilateral $O_1O_2O_3O_4$ is a parallelogram because the perpendicular bisectors of the sides of the triangles intersect at right angles.
- The area of the parallelogram $O_1O_2O_3O_4$ can be calculated using the formula for the area of a parallelogram:
$$\text{Area} = \text{base} \times \text{height}$$
- The length of the sides of the parallelogram can be related to the diagonals $AC$ and $BD$ of the original quadrilateral $ABCD$:
$$O_1O_2 = O_3O_4 = \frac{BD}{2 \sin(60^\circ)} = \frac{BD}{\sqrt{3}}$$
$$O_1O_4 = O_2O_3 = \frac{AC}{2 \sin(60^\circ)} = \frac{AC}{\sqrt{3}}$$
- Therefore, the area of $O_1O_2O_3O_4$ is:
$$[O_1O_2O_3O_4] = \frac{AC}{\sqrt{3}} \times \frac{BD}{\sqrt{3}} = \frac{AC \times BD}{3}$$

3. Calculate the ratio of the areas:
- The ratio of the area of $ABCD$ to the area of $O_1O_2O_3O_4$ is:
$$\frac{[ABCD]}{[O_1O_2O_3O_4]} = \frac{\frac{AC \times BD \times \sqrt{3}}{4}}{\frac{AC \times BD}{3}} = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{3}} = \frac{\sqrt{3}}{4} \times 3 = \frac{3\sqrt{3}}{4} = \frac{3}{2}$$

The final answer is $\boxed{\frac{3}{2}}$