Circular Numbers

Given a positive integer $a$ , let $b$ be the integer formed by shifting the digits of $a$ one decimal place to the left (and moving the first digit to the last - i.e. $123$ becomes $231$ ) and $c$ the integer formed by shifting the digits of $b$ in a similar manner.

Find is the least integer $a$ such that: $\frac{b}{a} = \frac{c}{b} = \frac{3}{2}$