Let $a,$ $b,$ $c$ be a three-term arithmetic series where all the terms are positive, such that $abc = 64.$  Find the smallest possible value of $b.$
By AM-GM,
\[\frac{a + b + c}{3} \ge \sqrt[3]{abc} = \sqrt[3]{64} = 4.\]Since $a,$ $b,$ $c$ form an arithmetic series, $\frac{a + b + c}{3} = b,$ so $b \ge 4.$

Equality occurs when $a = b = c = 4,$ so the smallest possible value of $b$ is $\boxed{4}.$