The digital filter should have specification $\omega_r=\dfrac \pi2$

$$H(s)=\dfrac {S+0.1}{(s+0.1)^2+16}$$

Let $H(s)=\dfrac {S+a}{(s+a)^2+ b^2} $ where b is analog resonant frequency.

By comparing we get analog frequency $b = 4 = Ω_r$

Now,

$\Omega_r= \dfrac 2T\tan (\dfrac {w_r}2) \\ 4=\dfrac 2T\tan (\dfrac \pi4) \\ T=0.5$

By BLT, digital filter is given by

$H(z)=H(s)|s =\dfrac {2(2-1)}{T(2+1)}$

by putting $T=0.5$

$$H(z)=\dfrac {0.125+0.006z^{-1}-0.122z^{-2}}{1+0.0006z^{-1}-0.9752^{-2}}$$