With no delay, here is the interactive ISCO calculator. Grab either of the red circles on one of the two curves and drag. Horizontal axis is $a/M$, vertical axis is $r_{ISCO}/M$. Scroll down for an explanation.

Update 2015-10-08: After some suggestions I have created the Kerr Calculator v2.

Explanation

The ISCO is the Innermost Stable Circular Orbit. Here I am referring to massive test particles in the equatorial plane of the Kerr black hole. In the spherically-symmetric Schwarzschild spacetime, there is only one ISCO, at $r=6M$. However, in Kerr, rotation splits prograde and retrograde orbits, and the ISCO may range all the way from $1M$ out to $9M$ for respectively prograde and retrograde orbits in extremal ($a=M$) Kerr, and anywhere in between, depending on the value of the spin parameter.

More than once I’ve been discussing with a colleague and asked either “What spin has an ISCO of $r$?” or “If the ISCO is at $r$, what is $a$?” I can never answer immediately because the formula for the ISCO is not so simple. This page should solve this problem.

The two curves above come from two roots of a quartic polynomial (see the derivation in e.g. Chris Hirata’s notes). Recall that quartics can be solved by radicals! Turn the crank, find the two appropriate roots. The Kerr ISCO formula is: \[ \frac{r_{ISCO}}{M} = 3 + Z_2 \mp \sqrt{(3-Z_1)(3+Z_1+2Z_2)}. \] The smaller root is for prograde orbits, and the larger root is for retrograde orbits. Here \begin{align} Z_1 &= 1 + \left(1 - \chi^2\right)^{1/3} \left((1 + \chi)^{1/3} + (1 - \chi)^{1/3}\right),\\ Z_2 &= \sqrt{3 \chi^2 + Z_1^2}, \end{align} and $\chi\equiv a/M$ is the dimensionless spin parameter, $-1\le \chi \le 1$.

Thanks to Niels Warburton for suggesting the JSXGraph toolkit.