Once you get circumference $ = 20\pi$, just divide by $4$ to get the quarter circumference.
Answer is $20\pi / 4 = 5\pi$. simple.
However, the title of the question seems to be misleading. The question in the body seems to have nothing to do with the area and the circumference (perimeter?) of the quadrant.
![enter image description here](https://i.stack.imgur.com/q2kQ8.png)
Circle $A$, with radius $r$, gets back to its starting point when $A$'s centre completes one rotation (around the centre of circle $B$ with radius $R$). Clearly $A$'s centre traverses a circular path of radius = $R+r$.
Now, Physics to the rescue.
Let the speed of $A$'s centre = $v$
Let the angular speed of $A$'s rotation around its own centre = $\omega$
Since, $A$ rolls over $B$ $\implies$ $v = \omega r$ (Assuming $B$ is fixed, the condition of rolling is that the point of contact is at rest).
Let the time taken by $A$'s centre to complete one rotation be $t$.
Then, $2\pi (R+r) = vt$
$\implies t = \frac{2\pi (R+r)}{v}$
Total angular distance traversed by $A$ around its centre in the same duration: $\theta = \omega t$
Using the above results, we get $\theta = \frac{v}{r} \frac{2\pi (R+r)}{v} = \frac{2\pi (R+r)}{r}$
In this time $t$, $A$ completes, say, $N$ rotations around its centre.
$\implies N = \frac{\theta}{2\pi} = \frac{(R+r)}{r}$
Now, In your case $ r = \frac{R}{3}$
Using this information, $ N = 4$
Best Answer
Circle $C_1: x^2+y^2+4x+22y+l=0$ has its center $(-2, -11)$ & a radius $\sqrt{(-2)^2+(-11)^2-l}=\sqrt{125-l}$
Similarly, circle $C_2: x^2+y^2-2x+8y-m=0$ has its center $(1, -4)$ & a radius $\sqrt{(1)^2+(-4)^2-(-m)}=\sqrt{m+17}$
Now, solving the equations of circles $C_1$ & $C_2$ by substituting the value of $(x^2+y^2)$ from $C_2$ into $C_1$, we get the $\color{blue}{\text{equation of common chord}}$ as follows $$(2x-8y+m)+4x+22y+l=0$$ $$\color{blue}{6x+14y+(l+m)=0}\tag 1$$
Now, since the circumference of circle $C_2$ is bisected by the circle $C_1$ hence the center $(1, -4)$ of circle $C_2$ must lie on the common chord or in other words, the common chord must pass through the center of circle $C_2$
Now, satisfying the above equation of common chord by center point $(1, -4)$ as follows $$6(1)+14(-4)+(l+m)=0$$ $$6-56+l+m=0$$ $$\bbox[5px, border:2px solid #C0A000]{l+m=50}$$