Aqueous zinc oxide reacts with an excess NaOH, and the final products are:

Zinc hydroxide can be prepared by adding a solution of sodium hydroxide, but not in excess, to a solution of any zinc salt. A white precipitate appears:

${\mathrm{Zn}}^{2+}+2{\mathrm{OH}}^{-}\to \mathrm{Zn}{\left(\mathrm{OH}\right)}_2$

If excess sodium hydroxide is added, the zinc hydroxide precipitate dissolves to form a colorless solution of zinc ions:

$\mathrm{Zn}{\left(\mathrm{OH}\right)}_2+2{\mathrm{OH}}^{-}\to \mathrm{Zn}{\left(\mathrm{OH}\right)}_4^{2-}$

Sodium zincate solutions can be prepared by dissolving zinc, zinc hydroxide or zinc oxide in an aqueous solution of sodium hydroxide.

So we can say that zinc oxide reacts with excess sodium hydroxide to form sodium zincate and water. 

$\mathrm{ZnO}+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{ZnO}}_2+{\mathrm{H}}_2\mathrm{O}$

So we can conclude that option A is the correct answer to this question.