Answer
$r=110 \mathrm{~mm}$
Work Step by Step
To find the one that is greater than $R$, we solve
$$
\frac{\mu_0 \varepsilon_0 R^2}{2 r} \frac{d E}{d t}\\=\frac{\mu_0 \varepsilon_0 R}{4} \frac{d E}{d t}
$$
for $r$. The result is $r=2 R=2(55.0 \mathrm{~mm})\\=110 \mathrm{~mm}$.
