How to normalize predesigned IIR filter coefficients to have unity gain?
In general, divide each of the $b$ coefficients by the maximum gain of your filter. This is equivalent to adding an attenuator before or after the original filter.

In this particular case, the maximum gain seems to occur at frequency 0 (where there is a double zero-pole cancellation).

At frequency 0, the gain of your filter is 36. So divide $b[]$ by 36.

Edit: the math

The frequency response comes from evaluating $H(z)$ at $z = e^{j\theta}$. Substituting and operating, and ignoring the phase, you get:

$|H(e^{j\theta})| = \frac{\sin(\theta/2)^2}{\sin(6\theta/2)^2}$

Taking the limit as $\theta \rightarrow 0$ you get $H(e^{j0}) = 36$.

Note that this is a simpler (and well known) filter applied twice. It corresponds the moving average of 6 samples (with the 1/6 factor missing).

So it is like two filters in cascade, each of which sums 6 consecutive samples. Each is a low-pass filter with gain 6 at 0 frequency.