How to determine the iron content in spinach with a spectrometer?
Here's how I approached the problem.

Plot the Data

Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)



The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.

Calculate the Concentration of Fe Ions Present

You have measured an absorbance of 0.015 with a corresponding concentration of $c_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.

The mass of iron in the solution is given by
$$ m_\ce{Fe} = c_\text{meas} \cdot M_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$
with the volume $V=0.02~\mathrm{L}$ and molar mass $M_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.

Calculate Mass Percentage

The mass of iron in one gram of spinach is easily calculated:
$$ w_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$

(I could just calculate away the units, but I was too lazy to do it here.)

The mass in $100~\mathrm{g}$ of spinach follows accordingly:
$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$