I am looking for the area of the white region interior to a set of circles with radius A, oriented on the edge of a larger circle with radius B, spaced apart from each other with distance C. You can assume that C is less than 2 times A, so that each smaller circle overlaps with its neighbors. To clarify, I am looking for the area of the circle with radius b, not covered by small circles.
On
Below is an example figure with $n=8$ small circles
Given $n$ small circles, the white area can be determined as the area of the large circle with radius $B$ minus the areas of the $n$ green sectors with angle $\alpha_1$ and the $n$ blue sectors with angle $\alpha_2$, i.e.: $$\text {Area} = \pi B^2- nA^2\frac{(\alpha_1 + \alpha_2)}{2}$$ The problem now is to find $\alpha_1$ and $\alpha_2$. Finding $\alpha_1$ is easy using the Cosine Rule, as we have an isosceles triangle with sides $A$ and base $C$, which gives $$\cos \alpha_1 = 1- \frac{C^2}{2A^2}$$
Finding $\alpha_2$ is harder. But we can use the triangle with the angle $\alpha_3$ shown above. We know that $2\alpha_3 = \frac{2\pi}{n}$ and hence $\alpha_3 = \frac{\pi}{n}$. We also know two sides of the triangle and can therefore use the Sine Rule to find $\alpha_4$: $$\frac{\sin \alpha_3}{A}=\frac{\sin \alpha_4}{B}$$ which gives $$\sin \alpha_4= \frac{B}{A} \sin \alpha_3$$ Hence $$\alpha_4 =\pi-\text {arcsin} (\frac{B}{A} \sin \alpha_3)$$ and therefore $$\alpha_2 = 2(\pi-\alpha_3-\alpha_4)$$
I suspect there is a simpler answer to this question, but I couldn't immediately find it.
It's more convenient to use the number of circles $n$ instead of the distance between their centers.
Then \begin{align} \phi&=\tfrac\pi{n} , \end{align} the distance between the centers $|C_iC_{i+1}|=2\,R\sin\tfrac\phi2$ and the total area $S$ of the interior region consists of $n$ areas $S_p$ of petals $P_1OP_2$.
\begin{align} S&=n\,S_p ,\\ S_p &=2\,S_{\triangle OC_0P_1}-S_{\mathrm{seg}\,C_0P_1P_2} \\ &=|OC_0|\cdot|P_1F|-\tfrac\theta2\,|C_0P_1|^2 \\ &=R\cdot r_a\sin\tfrac\phi2-\tfrac\theta2\,r^2 \tag{1}\label{1} . \end{align}
In $\triangle P_1OF$, $\triangle P_1FC_0$, $\triangle P_1C_0G$
\begin{align} \angle C_0GO&=\tfrac\pi2 ,\\ \angle GOC_0&=\tfrac\phi2=\tfrac\pi{n} ,\\ \angle FP_1O&=\tfrac\pi2-\tfrac\phi2 ,\\ \angle FC_0P_1&=\tfrac\theta2 ,\\ \angle C_0P_1F&=\tfrac\pi2-\tfrac\theta2 ,\\ \angle GP_1C_0& =\tfrac\phi2+\tfrac\theta2 =\tfrac\pi{n}+\tfrac\theta2 ,\\ |C_0G|&=R\sin\tfrac\phi2=R\sin\tfrac\pi{n} , \end{align}
\begin{align} \sin\angle GP_1C_0&= \sin(\tfrac\pi{n}+\tfrac\theta2) =\tfrac{|C_0G|}{r} =\tfrac{R}r\,\sin\tfrac\pi{n} ,\\ \tfrac\theta2&= \arcsin\left(\tfrac{R}r\,\sin\tfrac\pi{n}\right)-\tfrac\pi{n} ,\\ \sin\tfrac\theta2&= \tfrac{R}{r}\,\sin\tfrac\pi{n}\, \left( \cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}} \right) ,\\ r_a&=\frac{r}{\sin\tfrac\pi{n}}\,\sin\tfrac\theta2 \\ &= R\, \left( \cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}} \right) . \end{align}
Finally,
\begin{align} S(n,R,r)&= \pi\,r^2+ n\cdot \left( R^2\sin(\tfrac\pi{n})\, \left( \cos(\tfrac\pi{n})-\sqrt{\tfrac{r^2}{R^2}-\sin^2(\tfrac\pi{n})} \right) -r^2\, \arcsin\left(\tfrac{R}r\,\sin(\tfrac\pi{n})\right) \right) . \end{align}
Edit An example with (more-or-less) nice expression for the area.
Let $n=6$, $R=\sqrt3$, $r=1$. Then
\begin{align} r_a&=\sqrt3\left(\tfrac{\sqrt3}2-\sqrt{\tfrac13-\tfrac14} \right) =\tfrac32-\sqrt{1-\tfrac34}=1 ,\\ \tfrac\theta2&=\arcsin(\tfrac{\sqrt3}2)-\tfrac\pi6=\tfrac\pi6 ,\\ S(6,\sqrt3,1) &= \pi+6\cdot \left( \tfrac32\, \left( \tfrac{\sqrt3}2 -\sqrt{\tfrac{1}{3}-\tfrac{1}{4}} \right) -1\cdot \arcsin\tfrac{\sqrt3}2 \right) \\ &= \pi+ 9\, \left( \tfrac{\sqrt3}2 -\tfrac{\sqrt3}6 \right) -6\cdot \tfrac\pi3 \\ &= 3\sqrt3-\pi\approx 2.05 . \end{align}
In the image below four grid cells represent one square unit: