A simple calculation

To make a rhombicuboctahedron with unit sides, start with a unit cube, and on each its $6$ faces attach a $1 \times 1 \times \frac{\sqrt{2}}{2}$ box, and along each of its $12$ edges attach a triangular prism with a height of $1$ and legs of $\frac{\sqrt{2}}{2}$ , and along each of its $8$ vertices attach a right tetrahedron with legs of $\frac{\sqrt{2}}{2}$ . The volume is then:

$V_{\text{rhomb}} = V_{\text{cube}} + 6 \cdot V_{\text{box}} + 12 \cdot V_{\text{prism}} + 8 \cdot V_{\text{tetra}} = 1^3 + 6 \cdot 1 \cdot 1 \cdot \frac{\sqrt{2}}{2} + 12 \cdot \frac{1}{2} \cdot \bigg(\frac{\sqrt{2}}{2}\bigg)^2 \cdot 1 + 8 \cdot \frac{1}{6} \cdot \bigg(\frac{\sqrt{2}}{2}\bigg)^3 = 4 + \frac{10}{3}\sqrt{3}$

Therefore, $a = 4$ , $b = 10$ , $c = 3$ , and $a + b + c = \boxed{17}$ .