Answer
$1$
Work Step by Step
$tan^{-1}x^{2}=\Sigma_{n=0}^\infty(-1)^{n}\frac{(x^{2})^{2n+1}}{2n+1}$
$=\Sigma_{n=0}^\infty(-1)^{n}\frac{x^{4n+2}}{2n+1}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{4n+6}}{2n+3}}{\frac{x^{4n+2}}{2n+1}}|$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{x^{4}(2+1/n)}{(2+3/n)}|$
$=|\frac{x^{4}(2+0)}{(2+0)}|$
$=|x^{4}|$
This converges when $|x^{4}|\lt 1$ or $=|x| \lt 1$
Thus, the series has radius of convergence $1$.