What ion remains in the largest concentration at equilibrium: 0.1 M H₃PO₄ and 0.2 M KOH?
Actually, the ion with the largest concentration is $\ce {K+}$. 

But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.

You can find the exact concentrations if you need.

Addendum:
Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$
Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.  

Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.

Calculations:
Find $\ce {[H_2PO_4^{-}]}$:
$$\ce {H+ + HPO4^{2-} &lt;=&gt; H_2PO_4^{-}}$$
$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$
$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$
$[H_2PO_4^{-}]=3.75\times 10^{-10}$M  


Find $\ce {[PO_4^{3-}]}$:
$$\ce {HPO4^{2-} &lt;=&gt; H^+ + PO_4^{3-}}$$
$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$
$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$
$[PO_4^{3-}]=2.25\times 10^{-7}$