Answer
Products : $HNO_2$ and $OH^-$.
The left side will be favored.
Work Step by Step
$H_2O$ will give one proton to $N{O_2}^-$.
$N{O_2}^-(aq) + H_2O(l) HNO_2(aq) + OH^-(aq)$
Since $HNO_2$$(Ka = 4.5 \times 10^{-4})$ is stronger than $N{O_2}^-$$(Kb = 2.2 \times 10^{-11})$, the left side will be favored.