Answer: $290 \;km/s, 44.5^{\circ}$

The horizontal and vertical components of the velocity of the plane are: $300\cos 30^{\circ}$ and $300 \sin 30^{\circ}$

The horizontal and vertical components of the velocity of the wind are: $-80\cos 45^{\circ}$ and $80\sin 45^{\circ}$

Therefore the resultant velocity of the airplane is:

Horizontal component: $300\cos 30^{\circ}$ - $80\cos 45^{\circ} = 203.2$

Vertical component: $300 \sin 30^{\circ}$ + $80\sin 45^{\circ} = 206.6$

$R = 203.2 \hat{i} + 206.6 \hat{j}$

$\Rightarrow$ Magnitude of velocity $ = \sqrt{ 203.2^2 + 206.6^2} = 290\;km/s$

$\Rightarrow$ New Direction/Bearing of the airplane $ =\tan^{-} \large\frac{206.6}{203.3}$$ =44.5^{\circ}$