What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?
First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:

$$
\begin{align}
\ce{NaNO3 &amp;&lt;=&gt;&gt; Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &amp;&lt;=&gt;&gt; Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$

Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.

As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):

$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$

According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):

$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$

$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$

Finally, we can plug all the known quantities back to \eqref{eqn:1}:

$$
\begin{align}
c(\ce{NO3-}) &amp;= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
  &amp;= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
  &amp;= \pu{1.35 mol L-1}
\end{align}
$$