Cascaded one-pole sections and relationship to time constant
The difference equation of your single pole filter is

$$y[n]=x[n]+b(y[n-1]-x[n])\tag{1}$$

where $n$ is the time (sample) index, $x[n]$ is the input, and $y[n]$ is the output. Its transfer function is

$$H(z)=\frac{1-b}{1-bz^{-1}}\tag{2}$$

If you cascade $N$ identical single pole filters, the total transfer function is

$$G(z)=H^N(z)=\frac{(1-b)^N}{(1-bz^{-1})^N}\tag{3}$$

The inverse $\mathcal{Z}$-transform of $(3)$ is the impulse response of the total system:

$$g[n]=(1-b)^N\frac{(n+1)(n+2)\ldots (n+N-1)}{(N-1)!}b^n,\quad n\ge 0,\quad N&gt;1\tag{4}$$

From $(4)$ it's straightforward to compute the desired value of $b$ for $g[n]$ to decay to $0.01$ of its initial value in a given time $T$. The number of sample values corresponding to time $T$ is $K=\text{round}(Tf_s)$, where $f_s$ is the sampling frequency. We now need to solve

$$g[K]=0.01\cdot g[0]$$

for $b$, which results in

$$b=\left(\frac{0.01\cdot (N-1)!}{(K+1)(K+2)\ldots (K+N-1)}\right)^{1/K}\tag{5}$$

For $K=5$, as in your example, and for $N=2$ cascaded filters you get from $(5)$ a value of $b=0.27821$. For $N=5$ you would need to decrease this value to $b=0.15133$.