We can think of the more general case of n coins A of radius $a$ and a coin B of radius $b$.
Firstly, we note that the chain encloses a regular polygon of n sides, since they must be tangentially touching. Therefore, the angle between a coin and its neighbours is given by the formula for the internal angles of a regular polygon, i.e.
$$ \theta = \frac{360º}{n}$$
This angle determines the length $L_\theta$ that a single point would be able to cover on each A coin:
$$L_\theta = \frac{\theta}{360º} 2\pi a$$
However, what we must find is the "walkable" length $L_W$ that coin B is able to cover on each A coin. Then, the total length covered could be found as $L_T=nL_W$ and the number of rotations of B could be obtained as $R=\frac{L_T}{2\pi b}$
The length $L_W$ can be calculated by considering the situation in which coin B is tangentially touching two A coins, since that is what restricts its motion. In that case, a triangle is formed between their centers such that each angle between the center of B and the center of the other A coin is given by
$$ \beta = cos^{-1}(\frac{a}{a+b})$$
From this, we just subtract twice the associated length $L_\beta=\frac{\beta}{360º}2\pi a$ from $L_\theta$ to find $L_W$, since the restriction applies to each covered A coin twice.
$$ L_W=L_{\theta}-2L_{\beta}$$
Therefore,
$$ L_T = nL_W = n(L_{\theta}-2L_{\beta}) = n\left(\frac{\theta-2\beta}{360º}2\pi a\right)=\\
=2\pi an\left(\frac{\frac{360º}{n}-2cos^{-1}(\frac{a}{a+b})}{360º}\right)=\\
=2\pi a - \frac{4\pi a n \ cos^{-1}(\frac{a}{a+b})}{360º} $$
and
$$R(a,b,n)=\frac{2\pi a - \frac{4\pi a n \ cos^{-1}(\frac{a}{a+b})}{360º}}{2\pi b} = \frac{a}{b} \left( 1-n\frac{cos^{-1}(\frac{a}{a+b})}{180º} \right) $$
We can then just plug your numbers into the equation to find
$$R(2,1,10) = \frac{2}{1} \left( 10 \frac{cos^{-1}(\frac{2}{3})}{180º} \right) = 2\times 10 \times \frac{48.19º}{180º} = 20\times 0.267 = 5.35$$
So it rotates approximately 5.35 times in your case.
It should be noted that this assumes $n\geq 3$.
