= Solution
The <Fourier transform of a Gaussian> gives
$$
\mathcal F[e^{-t^2}]=\sqrt\pi e^{-\omega^2/4},\qquad\mathcal F^{-1}[2e^{-\omega^2/4}]=\frac2{\sqrt\pi}e^{-t^2}\equiv q(t).
$$
The <Fourier transform of a derivative> obeys $\mathcal F[q'']=-\omega^2\widetilde q$, so multiplication of the spectrum by $1+\epsilon\omega^2$ corresponds to $q-\epsilon q''$. Since $q''=(2/\sqrt\pi)(4t^2-2)e^{-t^2}$,
$$
\boxed{s(t)=\frac2{\sqrt\pi}\left(1+2\epsilon-4\epsilon t^2\right)e^{-t^2}.}
$$
This is real and even, as required by the given spectrum. Its time <integral> is two for every $\epsilon$, consistently with $\widetilde s(0)=2$.
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