In an AC resistive circuit, current and voltage are in phase.

So, $$I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)$$ ..... (i)

$$\therefore$$ Time period of one complete cycle of current is

$$T = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s$$

So, current reaches its maximum value at

$${t_1} = {T \over 4} = {1 \over {200}}s$$

When current is half of its maximum value, then from Eq.(i), we have

$$I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})$$

$$ \Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}$$

So, instantaneous time at which current is half of maximum value is $${t_2} = {1 \over {120}}s$$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$$\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3$$ ms