Although the specific circuit diagram is missing from the question text, this relies on a standard NEET 2016 problem. In the corresponding circuit, point A is at a potential of $+4 \text{ V}$ and is connected to the p-side of the diode, while point B is at $-6 \text{ V}$ and connected to the n-side through a $1 \text{ k}\Omega$ ($1000 \text{ } \Omega$) resistor. Since the p-side is at a higher potential than the n-side ($V_A > V_B$), the diode is forward-biased. An ideal diode offers zero resistance in forward bias. The potential difference across the resistor is $V = V_A - V_B = 4 \text{ V} - (-6 \text{ V}) = 10 \text{ V}$. According to Ohm's law, the current is $I = \frac{V}{R} = \frac{10 \text{ V}}{1000 \text{ } \Omega} = 0.01 \text{ A} = 10^{-2} \text{ A}$.