The concentration you have calculated is % ($w/w$). In other words, what that means is mass of solute in $\pu{100 g}$ of solution. In your case, it is mass of $\ce{NaCl}$ in $\pu{100 g}$ of solution: $\frac{12.27}{(12.27+14.23)}\times 100=46.3$%. However, in your statement, you have declared it is % ($w/v$), which is not.
the others's answer is also not quite correct, because they have assumed when $\ce{NaCl}$ is dissolved in certain volume of water, the volume stays same (won't change). If you are allowed to assume that, then, you can conclude (like others) that when $\pu{12.27 g}$ of $\ce{NaCl}$ is dissolved in $\pu{14.23 g}$ of $\ce{H2O}$ (or $\pu{14.23 mL}$ of $\ce{H2O}$ assuming density of water is $\pu{1.00 g/mL}$), the volume of solution is still $\pu{14.23 mL}$. Based on allowed assumption, the % $(w/v)$ concentration of your solution is: $\frac{12.27}{14.23}\times 100=86.2$% (the value close enough to your internet finding).
However, most important data missing in your question is the density ($\rho$) of the sought $\ce{NaCl}$ solution. That would give exactly the volume ($v$) of the solution since $v=m/\rho$. Thus, correct % $(w/v)$ concentration of your solution is: $\left(\frac{12.27}{\rho(12.27+14.23)}\times 100\right)$%.
