The equilibrium constant for this reaction is approximately ${10}^{-3}$.

${\mathrm{HPO}}_4^{-2}\left(\mathrm{aq}.\right)+{\mathrm{HCO}}_3^{-}\left(\mathrm{aq}.\right) \rightleftharpoons {\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}.\right)+{\mathrm{CO}}_3^{-2}\left(\mathrm{aq}.\right)$

Which is the strongest  conjugate base in this reaction? 

Conjugate base of weak acid is strong base.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{eq}}=\frac{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]\left[{\mathrm{CO}}_3^{-2}\right]}{\left[{\mathrm{HPO}}_4^{-2}\right]\left[{\mathrm{HCO}}_3^{-}\right]}={10}^{-3} \\ \underset{\mathrm{W}.\mathrm{Base}}{{\mathrm{HPO}}_4^{-2}}+\underset{ }{\underset{\mathrm{W}.\mathrm{Acid}}{{\mathrm{HCO}}_3^{-}}\rightleftharpoons \underset{\mathrm{S}.\mathrm{Acid}}{{\mathrm{H}}_2{\mathrm{PO}}_4^{-}}+\underset{\mathrm{S}.\mathrm{Base}}{{\mathrm{CO}}_3^{-2}}} \\ \mathrm{Since} \mathrm{the} {\mathrm{K}}_{\mathrm{eq}} \mathrm{is} \mathrm{very} \mathrm{less}, \mathrm{equilibrium} \mathrm{lies} \mathrm{on} \mathrm{the} \mathrm{L}.\mathrm{H}.\mathrm{S} \\ \mathrm{Equilibrium} \mathrm{always} \mathrm{moves} \mathrm{from} \mathrm{strong} \mathrm{acid} \mathrm{to} \mathrm{weak} \mathrm{acid}. \\ \mathrm{Thus} {\mathrm{H}}_2{\mathrm{PO}}_4^{-} \mathrm{is} \mathrm{the} \mathrm{strong} \mathrm{acid} \mathrm{and} {\mathrm{CO}}_3^{-2} \mathrm{is} \mathrm{the} \mathrm{strong} \mathrm{base}. \end{gathered}$$