Answer
$$\varepsilon=2\beta \pi r_0^2 B e^{-2\beta t} $$
Work Step by Step
We have here a loop that is shrinking and a constant flux, so we must have an induced emf.
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|=IR$$
where $\phi =AB\cos\theta$ and here $A$ is changing and $B$ is constant. Also $\theta=0^\circ$, so that
$$ \varepsilon=B\left|\dfrac{dA}{dt}\right|$$
$$ \varepsilon=B\left|\dfrac{d(\pi r^2)}{dt}\right|$$
where $r$ is given by $r=r_0e^{-\beta t}$
$$ \varepsilon=\pi B\left|\dfrac{d(r_0e^{-\beta t})^2}{dt}\right|$$
$$ \varepsilon=\pi r_0^2 B\left|\dfrac{d }{dt} e^{-2\beta t}\right|$$
$$ \varepsilon=\pi r_0^2 B\left| -2\beta e^{-2\beta t}\right|$$
$$\boxed{ \varepsilon=2\beta \pi r_0^2 B e^{-2\beta t} }$$
