Answer
$1.9\times 10^4Pa$
Work Step by Step
We can find the required pressure as follows:
First, we find the impulse for each molecule $J=2mv=2\times 28\times 1.6\times 10^{-27}\times 400=3.58\times 10^{-23}$
Now $F=5\times 10^{23}\times 3.58\times 10^{-23}=17.92N$
$p=\frac{F}{A}$
We plug in the known values to obtain:
$p=\frac{17.92}{10\times 10^{-4}}$
$\implies p=1.9\times 10^4Pa$