Computing the concentration of a reactant given formation of products
In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:

$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$

where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:

$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$

Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore

$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$

$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$

To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.