= Solution
Let $u(t)$ be the signed air speed along the straw towards the bubble. <Conservation of mass in a pipe> for the incompressible plug flow gives
$$
\pi\varepsilon^2u=\frac d{dt}\left(\frac{4\pi a^3}{3}\right),
\qquad
u=\frac{4a^2\dot a}{\varepsilon^2}.
$$
The unsteady <Bernoulli equation> along the straw, whose open end is at atmospheric pressure, gives
$$
p_{\rm in}-p_{\rm atm}=-\rho L\dot u=\frac{2\gamma}{a}
$$
by the <Young–Laplace equation>. Substitution yields
$$
a^3\ddot a+2a^2\dot a^2=-\frac{\gamma\varepsilon^2}{2\rho L}=-C.
$$
Set $w=a^2\dot a$. Since $a\dot w=-C$, the <chain rule> gives $w\,dw/da=-Ca$. The initial condition $\dot a(0)=0$ implies
$$
a^4\dot a^2=C(a_0^2-a^2),
\qquad
a^2\dot a=-\sqrt{C(a_0^2-a^2)}.
$$
Writing $a=a_0\sin\theta$ and integrating gives the implicit solution
$$
t=\frac{a_0^2}{2\sqrt C}\left[\frac\pi2-\theta+\sin\theta\cos\theta\right],
\qquad \theta=\sin^{-1}(a/a_0).
$$
At disappearance $a=0$ and $\theta=0$, so
$$
\boxed{t_*=\frac{\pi a_0^2}{2\varepsilon}\left(\frac{\rho L}{2\gamma}\right)^{1/2}.}
$$
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