Disk problem: This one is more mechanical, so let's do it first. Let $r=r(t)$ be the radius of one of its faces. Note that $r$ is changing. We are told how fast $r$ is changing. In symbols, what we are told amounts to
$$\frac{dr}{dt}=0.02.$$
We are asked how fast the area of a face (side, like a coin) is changing. Let $A=A(t)$ be the area of the face. We want to find out about $\dfrac{dA}{dt}$.
We are given the rate of change of something, and want to find the rate of change of something else. We therefore need a link between the two quantities. In our case, the link is through the familiar formula
$$A=\pi r^2.$$
Differentiate both sides with respect to $t$. To differentiate $r^2$, use the Chain Rule. We get
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$$
At the instant when $r=8.1$, we know everything on the right-hand side. At that instant, area is changing at the rate of $(2\pi)(8.1)(0.02)$ (square inches per second).
Plane problem: First step: Draw a diagram. Let $O$ be the position of the observer, and let $A$ be the point $1$ mile above the observer. Let $P$ be the position of the plane. The position $P$ is changing. Note that $\triangle OAP$ is right-angled.
Let $x=x(t)$ be the distance $OP$. This distance is changing. We know at what rate $x$ is changing: it is the speed of the plane. Do we know that
$$\frac{dx}{dt}=400.$$
We are asked how fast the distance of the plane from the observer is changing. So we are asked how fast $OP$ is changing. Let $OP=z$. The $z=z(t)$ is a function of $t$. We want to find out about
$$\frac{dz}{dt}.$$
We need a link between $x$ and $z$. In this case, that is provided by the Pythagorean Theorem. Since $OA=1$, we have
$$z^2=1^2+x^2.$$
Differentiate immediately* with respect to $t$, using the Chain Rule. Now you should be able to finish quickly. We get
$$2\frac{dz}{dt}=2x\frac{dx}{dt},\quad\text{or equivalently} \frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}.$$
Now freeze the situation $45$ seconds after the plane passed overhead. We can compute $x$ at that instant, and, using the Pythagorean Theorem, $z$, and now we know everything.
Note Many students would write instead that $z=\sqrt{1+x^2}$, and then differentiate. That's perfectly fine, a bit more work, a bit greater chance of error.
We use your notation. Let $c=c(t)$ be the distance from the plane to the radar station at time $t$. Let $b$ be the distance from the plane to the point where the climb began. Then
$$c^2=36+b^2-12b\cos \theta=36+b^2+12b\sin\phi,\tag{1}$$
where $\theta=125^\circ$ and $\phi=35^\circ$. Differentiating, we get
$$2c\frac{dc}{dt}=2b\frac{db}{dt}+12\sin\phi\frac{db}{dt}.\tag{2}$$
"Freeze" at time $6$ minutes after the climb began. We know everything on the right-hand side of (2), since at that instant $b=(6)(4)$.
We need to calculate $c$ at time $t=6$. Equation (1) does that, with $b=24$.
Best Answer
This is a related rates question. The general strategy:
Draw a picture. Invariably useful with these sort of problems. In the picture, label every quantity that is fixed, and every quantity that is changing with a name.
Write down the information you are given. Identify the rates you are told something about, and the rate you want to know something about.
Find an equation that relates the quantities whose rates you are being asked about. A relation among the quantities, not the rates.
Differentiate the equation you found in the previous step; this will give you an equation that relations the quantities and their rates of change.
Plug in all the information you have. Solve for the information you want to know.
Here, after you draw the picture, you'll see that it makes sense to think of the radar station as the origin, and the plane as flying on the line $y=3$. Let $p(t)$ be the horizontal position of the plane at time $t$ (so that the plane will be at the point $(p(t),3)$). Let $D(t)$ be the distance from the plane to the radar station.
You are told how $p(t)$ is changing: that is, you are given information about $\frac{dp}{dt}$. You are being asked about how the Distance is changing; that is, you are being asked to find $$\frac{dD}{dt}\Bigl|_{D=4}$$ (Actually, it's unclear if you want the derivative when $D$, the straight line from the plane to the radar, is $4$, or if you want it when $p(t)$ is $4$; I think it is the former, though).
So you wan to find some equation that relates $p$ and $D$. After you do that, taking the derivative of the equation with respect to $t$ will give you an equation that relates $p$, $D$, $\frac{dp}{dt}$, and $\frac{dD}{dt}$. You know the value of $\frac{dp}{dt}$, and you know the value of $D$ you want. So you should figure out what $p$ is for that $D$ (if necessary), plug everything in, and solve for $\frac{dD}{dt}$.