Answer
The bearing the pilot should fly is $358^{\circ}$ and the airspeed should be 169.4 mph
Work Step by Step
Let $a$ be the airspeed
Let $b = 11~mph$
Let $c$ be the resultant of the two vectors $a$ and $b$. Note that the magnitude of $c$ is the ground speed. Therefore $c = \frac{400~mi}{2.5~hr} = 160~mph$
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 $A = 328^{\circ}-180^{\circ} = 148^{\circ}$
We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{(11~mph)^2+(160~mph)^2-(2)(11~mph)(160~mph)~cos~148^{\circ}}$
$a = \sqrt{28706.13~(mph)^2}$
$a = 169.4~mph$
The airspeed should be 169.4 mph
We can use the law of sines to find angle $B$ which is the angle between vectors $a$ and $c$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$B = arcsin(\frac{b~sin~A}{a})$
$B = arcsin(\frac{(11)~sin~148^{\circ}}{169.4})$
$B = arcsin(0.03441)$
$B = 2.0^{\circ}$
The bearing the pilot should fly is $360^{\circ}-2.0^{\circ} = 358^{\circ}$