Given alternating emf

$E = 440\sin(100\pi t)$

Comparing with the standard form

$E = E_0\sin(\omega t)$

we get:

$E_0 = 440\text{ V}, \qquad \omega = 100\pi\text{ rad/s}$

Given inductance

$L = \frac{\sqrt{2}}{\pi}\text{ H}$

Inductive reactance

For a pure inductor,

$X_L = \omega L$

Substituting values:

$X_L = (100\pi)\left(\frac{\sqrt{2}}{\pi}\right) = 100\sqrt{2}\,\Omega$

Peak current

For a pure inductive circuit,

$I_0 = \frac{E_0}{X_L}$

So,

$I_0 = \frac{440}{100\sqrt{2}} = \frac{4.4}{\sqrt{2}}\text{ A}$

RMS current

An a.c. ammeter reads the rms current:

$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

Therefore,

$I_{\text{rms}} = \frac{4.4}{\sqrt{2}\cdot \sqrt{2}} = \frac{4.4}{2} = 2.2\text{ A}$

Match with options

$\boxed{2.2\text{ A}}$

So the correct option is C.