Shouldn't the side be a multiple of 6?

Let $GHIJKL$ be a hexagon formed by the intersection points of the circles inside the original hexagon. Let $R, S$ be midpoints of sides $AB$ and $BC$ respectively. By Pythagorean theorem we get $RG = RB \Rightarrow \angle RBG = 45 \Rightarrow$ by symetry $\angle GBH = 30$ . We can easily find $GH = \frac{7 \sqrt{3}-7}{2} \Rightarrow A(GHIJKL) = \frac{294 \sqrt{3} - 441} { 4}$ The shaded area is area of the hexagon $GHIJKL$ minus $6 \times$ area bounded by the circle and the side of the small hexagon. Those bounded areas can be calculated as $\triangle GBH -$ sector $GBH =$ $\frac{49 \pi - 147}{24}$ The shaded area is then: $\frac{7^2}{4}(6 \sqrt{3}-6-\pi)$ Final answer is $7+4+6+3+1=\boxed{21}$ .

$\angle RBG = 45$

Every circle intersects at 2 points on the Hexagon. Let M,N be 2 points on BC where BN > CN.

Now the area of arc $BHN$ = $\frac{45}{360}$ $\pi \frac{7^2}{2}$ = $\pi \frac{7^2}{16}$

Now triangle $BHC$ = $\frac{1}{2}$ * $\frac{7^2}{2}$

So the area of $MNH$ = $\pi \frac{7^2}{8}$ - $\frac{1}{2}$ $\frac{7^2}{2}$ = $\frac{7^2}{8}$ ( $\pi-2$ )

We have 6 section like $MNH$ . so total area = 6 $\frac{7^2}{8}$ *( $\pi-2$ )

Now the area of 6 half circles are = 6* $\frac{120}{360}$ $\pi \frac{7^2}{2}$ = $7^2\pi$

the area of the hexagon is = $\frac{7^2}{2}$ * $3\sqrt3$

the shaded area is = the area of the hexagon -( the area of 6 half circles -the area of 6 section like $MNH$ )

= $\frac{7^2}{2}$ $3\sqrt3$ - ( $7^2*\pi$ - 6 $\frac{7^2}{8}$ ( $\pi-2$ ))

= $\frac{7^2}{4}(6\sqrt3 - 6 -\pi)$

So. $7+4+6+3+1$ = $21$