Let $A$ be the position of the airplane, and let $B$ be the point on the road directly below the plane. Let $C$ be the location of the nearer kilometre marker, and let $D$ be the position of the further one. Make a suitable labelled diagram.
We are told that $\angle BAC$ is $30^\circ$ ($60^\circ$ _below_ the horizontal) and that $\angle BAD$ is $45^\circ$.
We have $$\frac{BC}{h}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$
Similarly, $$\frac{BD}{h}=\tan(45^\circ) =1.$$
Thus $BC=\frac{h}{\sqrt{3}}$ and $BD=h$. But $BD-BC=1$. This gives the equation $$h\left(1-\frac{1}{\sqrt{3}}\right)=1,$$ and now we can solve for $h$.