It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are:
$$ \begin{align}
\ce{As2O3 (s) -> H3AsO4 (aq)} \\
\ce{NO3- (aq) -> N2O3 (aq)}\\
\end{align}
$$
First balance $\ce{As}$ and $\ce{N}$ in both equations. Then, balance $\ce{O}$ in both equations with water and $\ce{H}$ with $\ce{H+}$:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq)}\\
\ce{2 NO3- (aq) + 6 H+ (aq) -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$
Finally, balance the charges by electrons:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\
\ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$
Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $\ce{e-}$s):
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$
Or simply,
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$