By Fermat's little theorem, we have $50^{82} \equiv 1 \pmod{83}$ because $83$ is prime and $(83, 50)=1$. $$50^{250} = 50^{246}\cdot 50^4 = \left(50^{82}\right)^3\times 2500^2 \equiv 1^3\times 10^2 = 100 \equiv \boxed{17} \pmod{83}.$$