calculate the number of shots of Captain Morgan’s rum
As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.

Take for instance this expression:

162 = A/(0.55 x 50) – (0.15 x 1)

Here, you have different units for each term; from the values you supplied, the units are

162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h

[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L]
If we use uniform units throughout, we get

1.62 g/L = A/27.5 L - 0.15 g/L

from where you get a more reasonable value for A:

A = (1.62 + 0.15) * 27.5 = 48.675 g

Each shot contains 

V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g

of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.

I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).