It looks like you've already started sketching the vector diagram in your image, which is a great start! Your sketch correctly shows the three displacement vectors and the resultant total displacement

Let East be the positive x-direction and North be the positive y-direction.

For $\mathbf{D_1} = 50 \text{ km East}$:

$$D_{1x} = 50 \text{ km}$$ $$D_{1y} = 0 \text{ km}$$

For $\mathbf{D_2} = 30 \text{ km North}$:

$$D_{2x} = 0 \text{ km}$$ $$D_{2y} = 30 \text{ km}$$

For $\mathbf{D_3} = 25 \text{ km at } 30^\circ \text{ East of North}$:
The angle $30^\circ$ is with respect to the North axis.

$$D_{3x} = 25 \sin(30^\circ) = 25 \times 0.5 = 12.5 \text{ km}$$ $$D_{3y} = 25 \cos(30^\circ) = 25 \times \frac{\sqrt{3}}{2} \approx 25 \times 0.8660 = 21.65 \text{ km}$$

Total East displacement ($D_x$):

$$D_x = D_{1x} + D_{2x} + D_{3x} = 50 + 0 + 12.5 = 62.5 \text{ km}$$

Total North displacement ($D_y$):

$$D_y = D_{1y} + D_{2y} + D_{3y} = 0 + 30 + 21.65 = 51.65 \text{ km}$$

The magnitude $D_{total}$ is found using the Pythagorean theorem:

$$D_{total} = \sqrt{D_x^2 + D_y^2}$$ $$D_{total} = \sqrt{(62.5)^2 + (51.65)^2}$$ $$D_{total} = \sqrt{3906.25 + 2667.7225}$$ $$D_{total} = \sqrt{6573.9725}$$ $$D_{total} \approx 81.08 \text{ km}$$

The direction $\theta$ is found using the arctangent function, measured from the East axis towards the North:

$$\theta = \arctan\left(\frac{D_y}{D_x}\right)$$ $$\theta = \arctan\left(\frac{51.65}{62.5}\right)$$ $$\theta = \arctan(0.8264)$$ $$\theta \approx 39.56^\circ$$

Since both $D_x$ and $D_y$ are positive, the direction is North of East.

The total displacement of the car is $\boxed{81.08 \text{ km at } 39.56^\circ \text{ North of East}}$.