6. Since $f(x)$ is an odd function, its graph is symmetric about the origin. Therefore, it is only necessary for the circle $x^{2}+y^{2}=R^{2}$ to pass through one of the highest points of $f(x)$. Let $\frac{\pi x}{R}=\frac{\pi}{2}$. Solving this, we find that the highest point of $f(x)$ closest to the origin is $P\left(\frac{R}{2}, 2 \sqrt{3}\right)$. By the problem statement,
$$
\left(\frac{R}{2}\right)^{2}+(2 \sqrt{3})^{2}=R^{2} \cdot R=4 \text {. }
$$

Thus, the smallest positive period of $f(x)$ is $\frac{2 \pi}{\frac{\pi}{R}}=8$. Therefore, the answer is A.
6. Given that on the graph of the function $f(x)=2 \sqrt{3} \sin \frac{\pi x}{R}$, an adjacent highest point and a lowest point exactly lie on $x^{2}+y^{2}=R^{2}$, then the smallest positive period of $f(x)$ is
A. 8
B. 4
C. 2
I. 1