Answer
The radius of convergence is $R=\dfrac{1}{e}$.
Work Step by Step
Root Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty} \sqrt[k] a_k$
1) When $0 \leq l \lt 1$. then the series $\Sigma a_k$ converges
2) When $l \gt 0$. 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 $a_k=(1+\dfrac{1}{k})^{k^2} \cdot x^k$
Now, $l=\lim\limits_{k \to \infty} \sqrt[k] {(1+\dfrac{1}{k})^{k^2} \cdot x^k}=|x| \lim\limits_{k \to \infty} (1+\dfrac{1}{k})^k=e|x|$
This means that the series converges for $r=e|x| \lt 1$ . Then $-1 \lt ex \lt 1 \implies \dfrac{-1}{e} \lt x \lt \dfrac{1}{e}$
So, the radius of convergence for the power series centered at $0$ is $R=\dfrac{1}{e}$
