$$x_{1}=2\text{ km; }y_1=0\text{ km}$$
Displacement for second leg of trip,
$$x_2=0\text{ km; }y_2=4.2\text{ km}$$
Adding these displacements gives the total displacement as,
$$\begin{split}\vec R &=\vec R_1+\vec R_2 \\\ &=(2\hat{i}+ 0\hat{ j})+(0\hat{i}+4.2\hat{j}) \\\ &=(2\hat{i}+ 4.2\hat{j})\text{ km} \\\ |\vec R| &=\sqrt{x^2_2+y^2_2}=\sqrt{2^2+4.2^2}\text{ km} \\\ &=\boxed{4.6 \text{ km}}\end{split}$$
To find the time the eagle is in the air, we can use the equation:
$$\text{speed}=\frac{\text{distance}}{\text{time}}$$
Since the eagle is flying at a constant speed, the average speed is given by:
$$v=\frac{\text{total distance}}{\text{total time}}$$
Solving for the total time and plugging in the average speed gives:
$$t=\frac{\text{total distance}}{\text{average speed}}=\frac{|\vec{R}|}{v}$$
Substituting the values we know, we get:
$$t=\frac{4.6\text{ km}}{1.5\text{ km/min}}=\boxed{3.1 \text{ min}}$$