Relative levels of the two most prominant glutamate species at pH 4.7
The two species are related. They are conjugated acid and base to each other. Assuming one species $\ce{A^2-}$ with concentration $x \,\pu{mol/L}$, the second species $\ce{HA-}$ with concentration $y\,\pu{mol/L}$. There is a third species $\ce{H2A}$ with concentration $z \pu{mol/L}$.
$$\frac{10^{-4.7}\times x}{y}=10^{-9.67} \tag{1}$$
$$\frac{10^{-4.7}\times y}{z}=10^{-2.19} \ \ \tag{2}$$
Based on above two equations, the two most prominant species are $\ce{HA-}$ and $\ce{H2A}$ and ratio should be $y/z$. I utilized definition of $K_\mathrm a$ instead of $\mathrm pK_\mathrm a$ to get the above two equations. $10^{-4.7}$ is simply $[\ce{H+}]$ at $\mathrm{pH} = 4.7$
Therefore the final answer is:
$$\frac{y}{z} = 10^{2.51}$$