Step 1: Express the pseudo steady state depletion rate from material balance:
For a closed reservoir under pseudo steady state, the volumetric material balance gives the reservoir average depletion rate as $\dfrac{dp}{dt} = -\dfrac{qB}{c_t V_p}$, where $q$ is the constant production rate, $B$ is the formation volume factor, $c_t$ is the total compressibility, and $V_p$ is the pore volume of the reservoir. Because $q$, $B$, $c_t$ and $V_p$ are all constant for the whole drainage volume, this depletion rate is a single constant value that applies uniformly to every point in the reservoir, not just to the wellbore.
Step 2: Recognize that the wellbore measurement represents the whole reservoir:
Since the same constant $dp/dt$ governs pressure decline everywhere under PSS, the 1.0 psi/day rate measured at the wellbore is identical to the rate of decline at the external boundary radius $r_e$. No additional adjustment is needed to convert a wellbore rate into a boundary rate under this flow regime.
Step 3: Calculate the cumulative pressure decline over the given time:
$\Delta p = 1.0 \times 500 = 500$ psi, using the constant depletion rate over the full 500 day period.
Step 4: Subtract from the known boundary pressure:
Given that the pressure at the external radius was $3500$ psi at the reference time, after $500$ days it becomes $3500 - 500 = 3000$ psi. This matches exactly what is predicted by the direct rate times time calculation, confirming the uniform depletion principle of pseudo steady state flow.
Final Answer:
\[\boxed{3000 \ \text{psi}}\]