$ω_r=\frac{π}{2}$

$H(s)= \frac{(s+0.1)}{((s+0.1)^2+16)}$ $H(s)= \frac{(s+0.1)}{((s+0.1)^2+(4)^2 )}$ From above equation we can say that, Ω=4. And The value of $ω_r$ is given as $ω_r=\frac{π}{2}$ Now we know that $Ω=\frac{2}{T_s} tan⁡(\frac{ω}{2})$ $∴4=\frac{2}{T_s} tan⁡(\frac{π}{4})$ $∴T_s=\frac{2}{4} tan⁡(\frac{π}{4})$ $∴T_s=0.5 sec$ Using bilinear transformation H(z) can be obtained by putting , $s=\frac{2}{T_s} \frac{((z-1)}{(z+1))}$in the equation of H(s) $∴H(z)=\frac{(4((\frac{(z-1)}{(z+1)}) )+0.1)}{([4(\frac{(z-1)}{(z+1)})+0.1]^2+16)}$ $∴H(z)=\frac{(\frac{(4z-4)}{(z+1)}+0.1)}{([\frac{(4z-4+0.1z+0.1)}{(z+1)}]^2+16 )}$ $∴H(z)=\frac{((4.1z-3.9)(z+1))}{(16.81z^2-31.98z+15.21+16(z+1)^2 )}$ $∴H(z)=\frac{(4.1z^2+4.1z-3.9z-3.9)}{(32.81z^2+0.02z+31.21)}$ $∴H(z)=\frac{(4.1z^2+0.2z-3.9)}{(32.81z^2+0.02z+31.21)}$