Solution: Given that the quadrature rule is exact for all polynomials of degree less than or equal to $3$. Note that for an odd function, the integral is zero and the right hand side expression is also zero, as $f(-a)=-f(a)$.

Take a function $f(x)=x^{2} $, as the degree is $2$, the integral must be exact. So we have, \begin{align*} & \int_{-1}^1 x^{2} ~\mathrm{d}x = f(-a)+f(a) \\ \implies & \left. \frac{x^3}{3}\right|_{-1}^ {1} = (-a)^2 + a^2 \\ \implies & \frac{1}{3}-\frac{-1}{3} = 2a^2 \\ \implies & \frac{2}{3} = 2a^2\\ \implies & a^2 = \frac{1}{3}. \end{align*} Thus, \[ \textcolor{blue}{\boxed{3a^2 = 1}} \]