How to determine the empirical formula given mass percentages of the elements and the molecular weight?
Rather than starting out with 100 g of unknown, I would start out with 1 mol of unknown, i.e. 283 g. That way, you can directly calculate how many moles of each element are in one mole of compound:

$$ n = 67.3 \%  \cdot \frac{\pu{283 g}}{\pu{12 g mol-1}} = \pu{15.86 mol}$$

Or to get the stoichiometric coefficient $\nu_C$ directly, take the percentage and multiply it by the ratio of compound molar mass to atomic mass:

$$ \nu_C =  67.3\% \cdot \frac{\pu{283 g mol-1}}{\pu{12 g mol-1}} = \pu{15.86}$$

If you do that for C, N, O and H, you get the formula:

$$\ce{C_{15.86}H_{19.4}N_{0.93}O_{3.80}}$$

You can try rounding this to the nearest integers:

$$\ce{C16H19NO4}$$

This compound has a molar mass of 293 g/mol. Playing around with the formula, you can find coefficients that match the molar mass exactly, e.g. $\ce{C16H13NO4}$ or $\ce{C17H17NO3}$. In any case, neither one of these nor the formula suggested by the OP matches both the molar mass and the mass percentages. 


  Is this correct?


$\ce{C17H21NO4}$ is a good answer, but the data are not consistent, so it is not the complete answer. I would say depending on whether the errors on the molar mass determination or the elemental composition are larger, you would have a different top candidate, and should report the numerically exact formula with non-integer coefficients. That way, the reader can critically evaluate the meaning of the data themselves.