Step 1: Convert the pore pressure into an equivalent mud weight instead of testing each option separately:
A useful shortcut in well control problems is to convert the given pore pressure directly into an equivalent circulating or static mud weight, sometimes called the pore pressure gradient in mud weight units, and then simply compare each candidate mud density against that single benchmark number rather than recomputing hydrostatic pressure four separate times.
Step 2: Compute the equivalent mud weight of the pore pressure:
Using the given conversion factor, $1$ lbm/gal of mud produces a gradient of $0.052$ psi/ft, so the pore pressure of $6500$ psig at a TVD of $10000$ ft corresponds to a gradient of \[ \text{gradient} = \frac{6500}{10000} = 0.65 \ \text{psi/ft} \] Converting this gradient into an equivalent mud weight by dividing by $0.052$ gives \[ \rho_{eq} = \frac{0.65}{0.052} = 12.5 \ \text{lbm/gal} \] This $12.5$ lbm/gal figure is the exact mud weight that would make the well perfectly balanced, so it becomes the pass or fail line for every option.
Step 3: Compare option A, 13.8 lbm/gal, against the benchmark:
Since $13.8 > 12.5$, this mud is heavier than the balance point and will overbalance the formation, keeping fluid from entering the well, so option A passes.
Step 4: Compare option B, 11.3 lbm/gal, against the benchmark:
Since $11.3 < 12.5$, this mud is lighter than the balance point and will underbalance the formation, allowing a kick, so option B fails.
Step 5: Compare option C, 11.8 lbm/gal, against the benchmark:
Since $11.8 < 12.5$, this mud is also lighter than the balance point, so it too underbalances the well and allows formation fluid to enter, so option C fails.
Step 6: Compare option D, 13.2 lbm/gal, against the benchmark:
Since $13.2 > 12.5$, this mud exceeds the balance point and keeps the well safely overbalanced, so option D passes.
Final Answer:
\[ \boxed{\text{Options A and D, 13.8 and 13.2 lbm/gal, are acceptable}} \]