Are my calculation correct?
$$\left(\frac{211}{307}\right)=-\left(\frac{307}{211}\right)=-\left(\frac{159}{211}\right)=-\left(\frac{53}{211}\right)-\left(\frac{3}{211}\right)=-\left(\frac{211}{53}\right)\times-\left(\frac{211}{3}\right)=\left(\frac{-1}{53}\right)\left(\frac{1}{3}\right)$$ (since 2 negatives cancel out).
Then knowing the facts that $\left(\frac{1}{3}\right)=1^{(3-1)/2}=1$ and $53 \equiv +-3 \pmod{8}\Rightarrow \left(\frac{-1}{53}\right)=-1$
Put everything together to get $-1$?
By Quadratic Reciprocity and the properties of the Legendre symbol:
$$\left(\frac{211}{307}\right)=-\left(\frac{307}{211}\right)=-\left(\frac{96}{211}\right)=-\left(\frac{2}{211}\right)^5\left(\frac{3}{211}\right)$$
$211\equiv 3\pmod{8}$ and $211\equiv 7\pmod{12}$, so by Quadratic Reciprocity $$=-(-1)^5(-1)=-1$$
As has been said on the Wikipedia page I linked, $\left(\frac{3}{p}\right)=1\iff p\equiv \pm 1\pmod{12}$.
Proof: If $p\equiv 1\pmod{4}$, then $\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=1\iff p\equiv 1\pmod{3}$, so $p\equiv 1\pmod{12}$ (see Chinese Remainder Theorem).
If $p\equiv -1\pmod{4}$, then $\left(\frac{3}{p}\right)=-\left(\frac{p}{3}\right)=1\iff p\equiv 2\pmod{3}$, so $p\equiv -1\pmod{12}$.