Yeah, I was able to hit 0.000285 by accounting for the assumed 4 Pa pressure difference.
You also missed a step in the first conversion, the infiltration rates are for per unit area, so we need to include the square foot to square meter conversion by multiplying 0.000019 m3/s by 10.7639 (sf in 1 m2) = 0.00205 m3/m2/s1
Now you can solve it using the following crack equation that relates pressure to flow rates through cracks:
\dot{Q} = \frac{1}{\rho} C \Delta{P}^n
C and n are unknown air flow coefficients, \rho is air density, \Delta{P} is the envelope pressure difference, and \dot{Q} is the resulting volume of air.
I approximated n to be from 0.6 to 0.7, and substitute 0.00205 m3 into the equation to solve for the 4 Pa pressure difference. \rho cancels out when you do the substitution and the resulting equation is:
\dot{Q} = 0.00205 \cdot \frac{4}{75}^n
You can see here for an n of around 0.67 we get the target 0.000285 m3/s/m2:
