Answer
The bearing the pilot should fly is $65^{\circ}30'$
The ground speed will be 180.9 mph
Work Step by Step
Let $a = 168~mph$
Let $b = 27.1~mph$
Let $c$ be the resultant of the two vectors $a$ and $b$. Note that the magnitude of $c$ is the ground speed.
The vectors a, b, and c form a triangle. Let angle $A$ be the angle that subtends side $a$. Then $A$ is the angle between vector b and vector c. Then $A = 57^{\circ}40'$
We can use the law of sines to find the angle B between the vectors a and c:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$B = arcsin(\frac{b~sin~A}{a})$
$B = arcsin(\frac{(27.1)~sin~57^{\circ}40'}{168})$
$B = arcsin(0.1363)$
$B = 7^{\circ}50'$
The bearing the pilot should fly is $57^{\circ}40'+7^{\circ}50' = 65^{\circ}30'$
We can find the angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-57^{\circ}40'-7^{\circ}50'$
$C = 114^{\circ}30'$
We can use the law of sines to find $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(168~mph)~sin~114^{\circ}30'}{sin~57^{\circ}40'}$
$c = 180.9~mph$
The ground speed will be 180.9 mph