Solve the Problem $\mathrm { BAC } = \frac { \text { ounces of alcohol } \times \text { percent alcohol } \times 0.075 } { \text { body weight in } \mathrm { lb } } - ($

Solve the problem.
-The Blood Alcohol Concentration (BAC) of a person who has been drinking is given by the formula $\mathrm { BAC } = \frac { \text { ounces of alcohol } \times \text { percent alcohol } \times 0.075 } { \text { body weight in } \mathrm { lb } } - ($ hours of drinking $\times 0.015 ) .$
For example, the BAC for a 150-lb man who in two hours has drunk two 12-ounce beers, each having a $3.3$ percent alcohol content is
$\mathrm { BAC } = \frac { ( 2 \times 12 ) \times 3.3 \times 0.075 } { 150 } = ( 2 \times 0.015 )$
Find the BAC for a 160-lb man who in three hours has drunk 3 12-ounce beers, each having a $3.5$ percent alcohol content.

A) $0.027$
B) $0.015$
C) $0.021$
D) $0.014$