An alternating voltage $$V(t) = 220\sin 100\pi t$$ volt is applied to a purely resistive load of $$50 \Omega$$. The time taken for the current to rise from half of the peak value to the peak value is:

$$V(t) = 220\sin(100\pi t)$$, $$R = 50\Omega$$.

Peak current: $$I_0 = V_0/R = 220/50 = 4.4$$ A.

Current: $$I = I_0\sin(100\pi t)$$.

Half peak: $$I_0/2$$ when $$\sin(100\pi t_1) = 1/2 \Rightarrow 100\pi t_1 = \pi/6 \Rightarrow t_1 = \frac{1}{600}$$ s.

Peak: when $$\sin(100\pi t_2) = 1 \Rightarrow 100\pi t_2 = \pi/2 \Rightarrow t_2 = \frac{1}{200}$$ s.

$$\Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} = \frac{3-1}{600} = \frac{2}{600} = \frac{1}{300}$$ s $$= 3.33$$ ms.

The answer is Option (2): $$\boxed{3.3 \text{ ms}}$$.