Calculate the pH of Na2CO3 Given information:

$K_\mathrm{a}(\ce{H2CO3}) = 4.3 \times 10^{-7}$,
$K_\mathrm{a}(\ce{HCO3-}) = 4.8 \times 10^{-11}$,
${{\ce{[Na2CO3]}=0.20\ \mathrm{M}}}$

This was a question in an exam and I'm struggling to understand how to solve it. I understand that $\ce{Na}$ is soluble in water, so $\ce{[CO3^{2-}]}=0.20\ \mathrm{M}$. I also know how $\ce{H2CO3}$ dissociates:

$$\ce{H2CO3 &lt;=&gt; H+ +HCO3- &lt;=&gt; 2H+ +CO3^{2-}}$$

When I try solve it using $K_\mathrm{a}=\frac{[P]}{[R]}$ I get:

$$\frac{\ce{[H+]}^2}{\ce{[H2CO3]}}=1.032 \times10^{-16}$$

Past here I'm not sure. I've been told to find $K_\mathrm{b}$ because it's a base but I'm not sure why that would make a difference. Please could someone make a worked solution or give me some hints?
It suffices to consider only the reaction

$$\ce{CO3^{2-} + H2O &lt;=&gt; HCO3- + OH-}$$

with $k_{B_1} = 2.1 \cdot 10^{-4}$, since the reaction

$$\ce{HCO3- + H2O &lt;=&gt; H2CO3 + OH-}$$

with $k_{B_2} = 2.3 \cdot 10^{-8}$ will only contribute a small amount of $\ce{OH-}$.

$$k_{B_1} = \frac{[\mathrm{{HCO_{3}}^{-}}] [\mathrm{OH^-}]}{[\mathrm{{CO_{3}}^{2-}}]} = \frac{x \cdot x}{c - x} \approx \frac{x \cdot x}{c}$$

$$x = [\mathrm{OH^-}] = \sqrt{k_{B_1} \cdot c} = 6.455 \cdot10^{-3}$$

which gives a pH of 11.8.