Answer
$d_{}=3.39 \times 10^{-3} \mathrm{~m}
$
Work Step by Step
Now the area plate be $A$ and the plate separation be $d$. The displacement current is
$$
i_d=\varepsilon_0 \frac{d \Phi_E}{d t}\\=\varepsilon_0 \frac{d}{d t}(A E)\\=\varepsilon_0 A \frac{d}{d t}\left(\frac{V}{d}\right)=\\\frac{\varepsilon_0 A}{d}\left(\frac{d V}{d t}\right) .
$$
Now the potential difference across the capacitor is the same in magnitude as the emf of the generator, so $V=\varepsilon_{\mathrm{m}} \sin \omega t$ and $d V / d t=\omega \varepsilon_{\mathrm{m}} \cos \omega t$. Thus, $i_d=\left(\varepsilon_0 A \omega \varepsilon_{\mathrm{m}} / d\right) \cos \omega t$ and $i_{d \max }=\varepsilon_0 A \omega \varepsilon_{\mathrm{m}} / d$. This means
$$
d=\frac{\varepsilon_0 A \omega \varepsilon_{\mathrm{m}}}{i_{d \max }}\\=\frac{\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right) \pi(0.180 \mathrm{~m})^2(130 \mathrm{rad} / \mathrm{s})(220 \mathrm{~V})}{7.60 \times 10^{-6} \mathrm{~A}}=3.39 \times 10^{-3} \mathrm{~m},
$$
where $A=\pi R^2$ was used.
