jmc

Converting $321_b$ to base 10 and setting it equal to 57, we find that $$\begin{array}{rl}3(b^2)+2(b^1)+1(b^0)& =57\\ 3b^2+2b+1& =57\\ \Rightarrow 3b^2+2b-56& =0\\ \Rightarrow (3b+14)(b-4)& =0\end{array}$$This tells us that $b$ is either $-\frac{14}{3}$ or $4$. We know that $b>0$, so $b=\boxed{4}$.