Given that $E=440\sin 100\pi t, L=\frac{\sqrt{2}}{\pi }\mathrm{H}$

Angular frequency of the source is $\omega =100\pi \mathrm{rad} {\mathrm{s}}^{-1}$.

Now the reactance of the inductor will be,

$X_L=\omega L=100\pi \frac{\sqrt{2}}{\pi }=100\sqrt{2} \mathrm{\Omega }$

Therefore, the peak current $I_0=\frac{E_0}{X_L}=\frac{440}{100\sqrt{2}}=2.2\sqrt{2} \mathrm{A}$

$AC$ ammeter reads RMS value therefore reading will be $I_{\mathrm{rms} }$

$I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=2.2 \mathrm{A}$