Answer
The radius of convergence for the power series centered at $0$ is $R'=
\dfrac{R}{a}$.
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=c_k (ax)^k$
Now, $l=\lim\limits_{n \to \infty}|\dfrac{c_{k+1} (ax)^{k+1}}{c_{k} (ax)^{k}}|=|ax| \lim\limits_{n \to \infty}|\dfrac{c_{k+1}}{c_k}|=|ax| \times \lim\limits_{n \to \infty} |\dfrac{1}{R}|$
This means that the series converges for $r=|ax| \dfrac{1}{R} \lt 1$ . Then $-1 \lt \dfrac{ax}{R} \cdot x \lt 1 \implies -\dfrac{R}{a} \lt x \lt \dfrac{R}{a}$.
So, the radius of convergence for the power series centered at $0$ is $R'=
\dfrac{R}{a}$
