Answer: 140

Solution:

All six hexagon vertices lie on $\Omega:$ $O_1$ and $O_2$ by hypothesis, and $A, B, C, D$ as intersection points with $\Omega.$ Let $R$ be the radius of $\Omega,$ and let the arcs cut off by the sides $AB,$ $BO_1,$ $O_1C,$ $CD,$ $DO_2,$ $O_2A$ be $2\alpha, 2\beta, 2\beta, 2\gamma, 2\delta, 2\delta$ (the two $\beta$'s because chords $BO_1 = O_1C = r_1,$ the radius of $\omega_1,$ and likewise $O_2A = O_2D = r_2$), so $\alpha + 2\beta + \gamma + 2\delta = \pi.$ Each chord equals $2R\sin(\text{half its arc}):$ $2R\sin\alpha = 2,$ $2R\sin\gamma = 16,$ and the chord $\overline{O_1O_2}$ subtends $2\beta + 2\gamma + 2\delta,$ giving $2R\sin(\alpha + \sigma) = 15$ where $\sigma = \beta + \delta.$ External tangency gives a second equation worth $15:$ $r_1 + r_2 = 2R(\sin\beta + \sin\delta) = 15.$

Since $\gamma = \pi - \alpha - 2\sigma,$ we have $\sin\gamma = \sin(\alpha + 2\sigma).$ Sum-to-product then gives $$\begin{aligned} 18 &= 2R\left[\sin(\alpha + 2\sigma) + \sin\alpha\right] \\ &= 4R\sin(\alpha + \sigma)\cos\sigma \\ &= 30\cos\sigma, \end{aligned}$$ so $\cos\sigma = \frac{3}{5},$ and similarly $14 = 4R\cos(\alpha + \sigma)\sin\sigma$ gives $R\cos(\alpha + \sigma) = \frac{35}{8}.$ Combining with $2R\sin(\alpha + \sigma) = 15$ yields $4R^2 = 225 + \frac{1225}{16},$ so $R^2 = \frac{4825}{64}.$ Also $15 = 2R(\sin\beta + \sin\delta)$ $= 4R\sin\frac{\sigma}{2}\cos\frac{\beta - \delta}{2}$ with $\sin\frac{\sigma}{2} = \frac{1}{\sqrt{5}},$ so $\cos\frac{\beta - \delta}{2} = \frac{15\sqrt{5}}{4R}$ and $\cos(\beta - \delta) = \frac{1125}{8R^2} - 1.$

Joining the center of $\Omega$ to the six vertices splits the hexagon into six triangles, so its area is $\frac{1}{2}R^2$ ${}\cdot \small \left[\sin 2\alpha + 2\sin 2\beta + 2\sin 2\delta + \sin 2\gamma\right].$ Now $R^2(\sin 2\beta + \sin 2\delta)$ $= 2R^2 \sin\sigma\cos(\beta - \delta)$ $= \frac{8}{5}\left(\frac{1125}{8} - R^2\right)$ $= \frac{835}{8},$ while $\sin 2\alpha + \sin 2\gamma$ $= -2\sin 2\sigma \cos\bigl(2(\alpha + \sigma)\bigr)$ $= -2 \cdot \frac{24}{25} \cdot \left(-\frac{95}{193}\right)$ since $\cos\bigl(2(\alpha+\sigma)\bigr) = 1 - \frac{2 \cdot 225}{4R^2}$ $= -\frac{95}{193},$ so $\frac{1}{2}R^2(\sin 2\alpha + \sin 2\gamma)$ $= \frac{1}{2} \cdot \frac{4825}{64} \cdot \frac{912}{965}$ $= \frac{285}{8}.$ The area is $\frac{835}{8} + \frac{285}{8} = 140.$