Unknown polygon!

Note $\overline{A_1A_k}$ is the base of an isosceles triangle with two sides of length $R$ and angle $\frac{2(k-1)\pi}{n}$ .

Then $\overline{A_1A_k}^2=R^2+R^2-2R\cdot R\cos\left(\frac{2(k-1)\pi}{n}\right)=2R^2\left(1-\cos\left(\frac{2(k-1)\pi}{n}\right)\right)$ . Therefore,

$\displaystyle\sum_{k=2}^n\overline{A_1A_k}^2=\sum_{k=2}^n2R^2\left(1-\cos\left(\frac{2(k-1)\pi}{n}\right)\right)$

$\displaystyle=2R^2\left(n-1-\sum_{k=2}^{n}\cos\left(\frac{2\pi (k-1)}{n}\right)\right)$ .

Recall that $1,\cos\left(\frac{2\pi}{n}\right),\cos\left(\frac{4\pi}{n}\right),\ldots,\cos\left(\frac{2(n-1)\pi}{n}\right)$ are the real parts of the $n$ -th roots of unity. Therefore, $\displaystyle\sum_{k=1}^n\cos\left(\frac{2\pi(k-1)}{n}\right)=0$ , so $\displaystyle\sum_{k=2}^n\cos\left(\frac{2\pi(k-1)}{n}\right)=-\cos(0)=-1$ .

Thus, $\displaystyle\sum_{k=2}^n\overline{A_1A_k}^2=2R^2(n-1-(-1))=2nR^2$ . Then since $2nR^2=14R^2$ , $n=\boxed{7}$ .

$For~a~n\!\!-\!\!sided~regular~polygon,~~angle~X=\dfrac{180^o} n.\\ A_1A_2, .......A_1A_r ...A_1A_n~~\text{ are all the bases of }\\ \text{isosceles triangles with equal sides R-R and 2*(r-1)*X as vertex angle.}\\ For~~A_1A_r, ~the~vertext~angle~is~~~=2*(r-1)*X.\\ \therefore~A_1A_r =2*R*Sin\{(r-1)X\}.\\ \implies~\{A_1A_r\}^2 =4*R^2Sin^2 \{(r-1)X \}.\\ \therefore for~ a ~n\!\!-\!\!gon~ \displaystyle \sum_{r=2}^n 4*R^2Sin^2\{(r-1)*X\}.\\For~n=5,~~X=\dfrac{180^o} 5=36^o .\\ \displaystyle \sum_{r=2}^5 4*R^2Sin^2\{(r-1)*36^o\}=10*R^2\\ For~n=7,~~X=\dfrac{180^o} 7 ,~~n-1=6.\\ \displaystyle \sum_{r=2}^7 4*R^2Sin^2\{(r-1)*\dfrac{180^o} 7\}=14*R^2 \\ n=~~~~\Large \color{#3D99F6}{7} \\~~\\$ $Intresting ~~observation.$ $\Large \color{#D61F06}{\displaystyle \sum_{r=2}^{n} \{A_1A_r\}^2=2*n*R^2}~~for~~n\!-\!gon.$