Because $f_X(x)$ is a density function, $\int_{-a}^a \frac{k}{a^2}(1-x^2)\,dx=1.$ (At the end we need to make sure that $|a|\le 1$, since a density cannot be negative.) Calculate. You will get a relationship between $a$ and $k$. As a check on your calculations, or mine, I think the relationship is $2k\left(a-\frac{a^3}{3}\right)=a^2.\tag{$1$}$
Then we find the mean. We could integrate. But the density function and the interval are symmetric about $x=0$, so the mean is $0$. It follows that since $\sigma^2=E(X^2)-(E(X))^2$, we have $\sigma^2=E(X^2)$. But $E(X^2)=\int_{-a}^a x^2\frac{k}{a^2}(1-x^2)\,dx.$ Integrate. You will get an expression for $\sigma^2$ in terms of $a$ and $k$. As a check on your calculations, the result is $2k\left(\frac{a^3}{3}-\frac{a^5}{5}\right)=a^2\sigma^2.\tag{$2$}$ Now you have two equations in two unknowns $a$ and $k$. Solve. Looks a bit messy. And it is. But if you use Equations $(1)$ and $(2)$ and divide, the $k$ disappears, and you arrive at a quadratic equation in $a^2$.