A

Step 1: Determine the parameters given in the problem:
Weight of piston, W = 21 lb
Deceleration, a = 2.1 ft/s²
Speed, v = 21 ft/s
Clearance, h = 0.001 in = 0.001/12 ft = 8.3333 x 10^{-5} ft

Step 2: Calculate the area of the piston:
- Diameter of piston, D = 5 in = 5/12 ft
- Radius, r = D/2 = (5/12)/2 ft = 5/24 ft
- Area, A = \pi r^2 = \pi \left(\frac{5}{24}\right)^2 = \pi \cdot \frac{25}{576} = \frac{25\pi}{576} ft²

Step 3: Apply the Newton's law and the equation for viscosity:
The drag force on the piston due to viscosity can be expressed as:
\[ F = \eta A \frac{dv}{dy} \]
Where:
- F = mass * acceleration = W/g (where g = 32.2 ft/s², gravitational acceleration)
- Viscosity, \( \eta \)
- dv/dy = v/h
\[ F = \eta A \frac{v}{h} \]
Substituting for F and rearranging gives:
\[ \eta = \frac{F h}{A v} \]

Step 4: Substitute values into the formula:
F = (21 lb) / (32.2 ft/s²) = 0.651 lb mass.
\[ \eta = \frac{(0.651 lb)(8.3333 \times 10^{-5} ft)}{ (\frac{25\pi}{576}) (21 ft/s)} \]
Step 5: Calculate \( \eta \):
\[ \eta \approx \frac{(0.651)(8.3333 \times 10^{-5})}{\frac{25\pi}{576} \cdot 21} \approx 0.001218 lb \, s/ft²\]

Therefore, the viscosity of the oil is approximately 0.001218 lb s/ft², which corresponds to option A.