A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA when it operates with a 6 V battery through a limiting resistor R. The value of R is:

400 Ω

As LED is connected to a battery through a resistance in series, hence the current floweing, is 10 mA (which is the same).

The voltage drop across LED = 2 V

As the battery has 6 V, the potential difference across R = 4 V

$∴ IR = 4 V$ or $R=\frac{4V}{10×10^{-3}A}=400 Ω$