Answer
$f$ is even.
Work Step by Step
To prove that a function is even, one must show that $$f(-x)=f(x) \,.$$
In this question
$$f(x)=a_{2n}x^{2n}+a_{2n-2}x^{2n-2} \cdots +a_2x^2+a_0=\sum_{i=0}^na_{2i}x^{2i} \, .$$
One can easily find $f(-x)$ by noting the fact that the function $f(x)$ is the sum of terms containing even powers of $x$ and for each $i=0, \cdots n$, one has $(-x)^{2i}=(-1)^{2i}x^{2i}=x^{2i}$.
So $$f(-x)=a_{2n}x^{2n}+a_{2n-2}x^{2n-2} \cdots +a_2x^2+a_0=\sum_{i=0}^na_{2i}x^{2i}=f(x) \, .$$
Hence, $f$ is even.
