Answer
$$f(x)
=x^2+ x^4+\frac{x^{6}}{2!}+\frac{x^{8}}{3!}+\frac{x^{10}}{4!}+....$$
Convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows
$$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
Now by comparison with the function $f(x)=x^2e^{ x^2}$, we have the Maclaurin series as follows
$$f(x)=x^2e^{ x^2}\\
=x^2(1+ x^2+\frac{x^{4}}{2!}+\frac{x^{6}}{3!}+\frac{x^{8}}{4!}+...)\\
=x^2+ x^4+\frac{x^{6}}{2!}+\frac{x^{8}}{3!}+\frac{x^{10}}{4!}+....$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)=x^2e^{ x^2}$ is also convergent for any value of $x$.
