Answer
$3\times10^{-3}kg/m^{3}$
Work Step by Step
Here we use Fick's law $\frac{m}{t}=\frac{DA\Delta C}{L}$ to find the lower concentration.
$\frac{m}{t}=\frac{DA\Delta C}{L}=\frac{DA(C_{2}-C_{1})}{L}$
$C_{1}=C_{2}-\frac{L(\frac{m}{t})}{DA}$ ; Let's plug known values into this equation.
$C_{1}=8.3\times10^{-3}kg/m^{3}-\frac{(0.02\space m)(4.2\times10^{-14}kg/s)}{(1.06\times10^{-9}m^{2}/s)(1.5\times10^{-4}m^{2})}$
$C_{1}=3\times10^{-3}kg/m^{3}$