A start: Draw the two diagonals of the square, and also the two lines that join midpoints of opposite sides. These $4$ lines divide the square into $8$ congruent right-angled isosceles triangles. By symmetry, it is enough to determine the probability that a randomly chosen point within a specific one of these triangles is closer to the centre of the square than it is to one of the sides of the square.
Assume (it makes no difference) that the original square has vertices $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$. Consider the triangle $T$ with vertices $(0,0)$, $(2,0)$, and $(2,2)$. This triangle has area $2$. So for our probability, we will determine the area of the region $K$ in $T$ consisting of all points $(x,y)$ such that the distance from $(x,y)$ to the origin is $\le$ the distance from $(x,y)$ to the nearest side of the original square. Then our required probability is the area of $K$ divided by $2$,
So let us determine the locus of all points $(x,y)$ in $T$ that are equidistant from the origin and the nearest side of the original square.
The distance to the origin is $\sqrt{x^2+y^2}$, and the distance to the nearest side of the original square is $2-x$. So our locus is $\sqrt{x^2+y^2}=2-x$. Square and simplify. We get $x=\frac{1}{4}(4-y^2)$, part of a parabola.
Our region $K$ is the region bounded by the $x$-axis, the parabola $x=\frac{1}{4}(4-y^2)$, and the line $y=x$. We find the area of $K$ by integrating with respect to $y$.
Setting up the integral is slightly tricky. We will be integrating from $y=0$ to where $x=\frac{1}{4}(4-y^2)$ meets $y=x$. That gives the equation $y=\frac{1}{4}(4-y^2)$, which has the solution $y=2\sqrt{2}-2$. Thus our required area is
$$\int_0^{2\sqrt{2}-2} \left(\frac{1}{4}(4-y^2)-y\right)\,dy.$$
After calculating the area, don't forget to divide by $2$.
Your process started out correct: The area of the board is
$$A_b=\pi r^2=81\pi$$
but the area of the bulls-eye is
$$A_e=\pi r^2=\pi$$
So the probability that any dart hits the bulls-eye is
$$P=\frac{A_e}{A_b}=\frac{\pi}{81\pi}=\frac{1}{81}$$
Now, to find the probability that exactly two darts hits the bulls-eye, first note there are exactly $\binom{4}{2}$ (read $4$ choose $2$) ways to choose exactly two darts from the four. This is defined as
$$\binom{4}{2}=\frac{4!}{2!(4-2)!}=\frac{24}{2\cdot 2}=6$$
Second, the probability that exactly two darts (in any order) hit the center is given by
$$P\cdot P\cdot (1-P)\cdot (1-P)=\frac{6400}{43046721}$$
Since there six ways to find arrange the darts and have this happen, the final probability of exactly two darts hitting the bulls-eye is given by
$$6\frac{6400}{43046721}=\frac{12800}{14348907}=0.0892054\%$$
Best Answer
So here are lots of details.
To get the area of the two inch rim, we want $$A = \pi(10)^2-\pi(10-2)^2 = \pi36.$$ But we are only interested in the section that is in the first quadrant, which is $$\frac{1}{4}A = \pi9.$$
So the probability of interest is $$\frac{\pi9}{\pi100} = \frac{9}{100}.$$
If you wanted to try to use complements, then if $R = \{\text{Hit 2 inch rim in Q1}\}$ and $\bar R = \{\text{Miss Q1 rim }\}$, then \begin{align*} P(R) &= 1-P(\bar R) \\ &=1-[P(\text{Hit quadrants 2,3,4})+P(\text{Miss rim in Q1 })]\\ &= 1-\left(\frac{3}{4}\left(\frac{\pi100}{\pi100}\right)+\frac{3}{4}\left(\frac{\pi(10-2)^2}{\pi(10)^2}\right)\right) \\ &=1-\frac{91}{100} \\ &= \frac{9}{100}. \end{align*}