A transfer function is called realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable.
The transfer function (as a function of $s$) is given by
$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$
The given frequency response is obtained form $(1)$ by substituting $s=j\omega$. Since $b>0$, $H(s)$ has two real-valued poles at $\pm\sqrt{b}$, and, consequently, the corresponding impulse response is two-sided (i.e., non-causal).
Partial fraction expansion of $(1)$ gives
$$H(s)=\frac{a}{2\sqrt{b}}\left[\frac{1}{\sqrt{b}-s}+\frac{1}{\sqrt{b}+s}\right]e^{-st_0},\qquad -\sqrt{b}<\textrm{Re}\{s\}<\sqrt{b}\tag{2}$$
From $(2)$ it is straightforward to obtain the impulse response
$$\begin{align}h(t)=\mathcal{L}^{-1}\{H(s)\}&=\frac{a}{2\sqrt{b}}\left[e^{\sqrt{b}(t-t_0)}u(-(t-t_0))+e^{-\sqrt{b}(t-t_0)}u(t-t_0)\right]\\&=\frac{a}{2\sqrt{b}}e^{-\sqrt{b}|t-t_0|}\tag{3}\end{align}$$
No matter how large you choose $t_0$, the impulse response will never be causal. However, since the system is stable, the impulse response decays for $|t|\to\infty$, so you can approximate the actual impulse response well by sufficiently shifting it to the right (i.e., by choosing some large $t_0$), and truncating the non-causal part.
Also take a look at this related question and its answer.
