What is the amount of nitrate ions in 20 g of Fe(NO3)3?
The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.

You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$

Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?


   There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.


What does that mean for the number of moles of nitrate ions?


   It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$


Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.


   $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$