NaN3 quantity for the generation of N2
$$\ce{2NaN3 -&gt; 2Na + 3N2}$$

As you already found the amount of $\ce{N2}$ required, I'll just give you a hint on how to get the amount of sodium azide required to produce that much nitrogen.

We know that 2 moles of sodium azide gives us 3 moles of nitrogen. So $x$ moles of sodium azide gives us $2$ moles of nitrogen. Solve for $x$. That's all!