First, when we pick 3 points inside the circle with radius $R$, 2 points are inside the smaller circle with radius same as radius of a remaining furthest point $(r_1, r_2 \le r_{\mathrm{furthest}} =q\lt R)$. And probablity for this condition expressed as $\frac{\pi q^2}{\pi R^2} \times \frac{\pi q^2}{\pi R^2} \times \frac{2\pi q \ dq}{\pi R^2} \times 3$. "$3$" corespond to each of the 3 points being furthest, and $2\pi q \ dq$ expresses area of annulus with radius $q$ and width $dq$ which containing $r_{\mathrm{furthest}}$. Since expected area of triangle with $(r_{\mathrm{furthest}} =q)$ is proportional to $q^2$ (this holds true not only for triangles), we can express it as $kq^2$, and we'll think about its true value later. Now, expected area of triangle that generate at random from inside the circle with radius $R$ expressed as $$\begin{split}E[S_\mathrm{all \ possible \ triangles}]&= \int_0^R (kq^2)(\frac{\pi q^2}{\pi R^2} \times \frac{\pi q^2}{\pi R^2} \times \frac{2\pi q \ dq}{\pi R^2} \times 3) \\ &=\frac{6k}{R^6}\int_0^R q^7 dq \\ &=\frac{3}{4} kR^2.\end{split}$$ Next, we think about value of $kq^2$; it is expected triangle area formed by 1 point $p_0$ is on a circumference with radius $q$, and other 2 points $p_1, p_2$ are inside the circle put randomly. In order for the expression doesn't have negative area, put the center of the circle (with radius $q$) at $(0,q)$. Then fix the $p_0$ at origin (it's on the rim of the circle there), and put other points $p_1$ and $p_2$ at $(r_1,\phi_1),(r_2,\phi_2)$. Now, triangle area that one point is fixed on the circumference $S_\mathrm{one \ point \ fixed}$ is expressed as $$S_\mathrm{one \ point \ fixed}=\left|\frac{1}{2}r_1 r_2 \sin (\phi_2 - \phi_1)\right| ,$$ and its probability $dP$ (probability that $p_1$ and $p_2$ is inside the integral area elements $dS_1$ and $dS_2$) is expressed as $$\begin{split}dP&=\frac{dS_1}{\pi q^2}\frac{dS_2}{\pi q^2} \\ &=\frac{r_1 dr_1 d\phi_1}{\pi q^2}\frac{r_2 dr_2 d\phi_2}{\pi q^2}.\end{split}$$ Then, integrate $S_\mathrm{one \ point \ fixed}dP$ over double circle domain ${D_\mathrm{circle}}^2$ give us the expected triangle area that one point is fixed on the circumference $kq^2$, where $D_\mathrm{circle}$ can be expressed as $(0\lt r\lt 2q\sin \phi),(0\lt \phi \lt \pi)$ because distance between origin and circumference is $2q\sin \phi$ (cf. Thales' theorem). Since the area must be positive, integrate $S_\mathrm{one \ point \ fixed}dP$ taking care to use the symmetry of $\phi_1\lt\phi_2$ and $\phi_2\lt\phi_1$. Also, using the interchange of the order of integration. $$\begin{align} kq^2&=\int_{{D_\mathrm{circle}}^2}S_\mathrm{one \ point \ fixed} \ dP \\ &=\int_{D_\mathrm{circle}}\int_{D_\mathrm{circle}}S_\mathrm{one \ point \ fixed}\frac{dS_1}{\pi q^2}\frac{dS_2}{\pi q^2} \\ &=\int_0^\pi \int_0^{2q\sin \phi_2} \int_0^\pi \int_0^{2q\sin \phi_1}\left|\frac{1}{2}r_1 r_2 \sin (\phi_2 - \phi_1)\right| \frac{r_1 dr_1 d\phi_1}{\pi q^2}\frac{r_2 dr_2 d\phi_2}{\pi q^2}\\ &=\boxed{2}\int_0^\pi \int_0^{2q\sin \phi_2} \int_0^{\boxed{\phi_2}} \int_0^{2q\sin \phi_1}\left(\frac{1}{2}r_1 r_2 \sin (\phi_2 - \phi_1)\right) \frac{r_1 dr_1 d\phi_1}{\pi q^2}\frac{r_2 dr_2 d\phi_2}{\pi q^2}\\ &=\frac{1}{\pi^2 q^4}\int_0^\pi \int_0^{\phi_2} \left(\int_0^{2q\sin \phi_2}{r_2}^2dr_2 \int_0^{2q\sin \phi_1}{r_1}^2dr_1 \right) \sin (\phi_2 - \phi_1) \ d\phi_1 d\phi_2 \\ &=\frac{1}{\pi^2 q^4}\int_0^\pi \int_0^{\phi_2} \frac{64q^6}{9}\sin^3 \phi_1 \sin^3 \phi_2 \sin (\phi_2 - \phi_1) \ d\phi_1 d\phi_2 \\ &=\frac{64q^2}{9\pi^2}\int_0^\pi \int_0^{\phi_2} \sin^3 \phi_1 \sin^3 \phi_2 (\sin \phi_2 \cos \phi_1 - \cos \phi_2 \sin \phi_1) \ d\phi_1 d\phi_2 \\ &=\frac{64q^2}{9\pi^2} \left\{\int_0^\pi \sin^4 \phi_2\int_0^{\phi_2} \sin^3 \phi_1 \cos \phi_1 \ d\phi_1 d\phi_2 -\boxed{\int_0^\pi} \sin^4 \phi_1 \boxed{\int_{\phi_1}^\pi} \sin^3 \phi_2 \cos \phi_2 \ \boxed{d\phi_2} \boxed{d\phi_1} \right\}\\ &=\frac{64q^2}{9\pi^2} \left\{\int_0^\pi \sin^4 \phi_2 \left(\frac{1}{4} \sin^4 \phi_2 \right) \ d\phi_2 -\int_0^\pi \sin^4 \phi_1 \left(-\frac{1}{4} \sin^4 \phi_1 \right) \ d\phi_1 \right\} \\ &=\frac{64q^2}{9\pi^2}\cdot\frac{1}{4} \left\{2\int_0^\frac{\pi}{2} \sin^8 \phi_2 \ d\phi_2 +2\int_0^\frac{\pi}{2} \sin^8 \phi_1 \ d\phi_1 \right\} \\ &=\frac{64q^2}{9\pi^2}\cdot\frac{1}{4}\cdot 2 \left(\frac{35\pi}{256}+\frac{35\pi}{256} \right) \ \ \ \ \ \ \ \ \ \ (\mathrm{Wallis' \ integrals}) \\ &=\frac{35\pi}{36}q^2.\end{align}$$ Finally, substitute $kq^2=\frac{35\pi}{36}q^2$ and $R=1$ for $E[S_\mathrm{all \ possible \ triangles}]$, we get the value we wanted to find: $$\begin{align}E[S_\mathrm{All \ possible \ triangles}]|_{R=1}&=\frac{3}{4}\cdot\frac{35\pi}{36}\cdot 1^2 \\ &=\frac{35\pi}{48}.\end{align}$$
