Answer
$$3$$
Work Step by Step
Given: $a_n=\dfrac{n!}{3 \cdot 6.....3n}x^n$
Let us apply the Ratio Test to the given series.
In order to solve this series, we have:
$\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=|\lim\limits_{n \to \infty} \dfrac{a_n (\dfrac{x}{3})}{a_n}| \\=\lim\limits_{n \to \infty} |\dfrac{x}{3}|$
Now, $|\dfrac{x}{3}| \lt 1$
and $|x| \lt 3$
The radius of convergence is: $3$
