Correct option is D. $$25\ mA$$
Since $$2000 \Omega$$ is parallel to Zener diode
So, the current passing through it as shown in the circuit,
$$i_{3} = \dfrac {50}{2000} = 25\ mA$$
Potential difference across $$1000\Omega$$,
$$V_{1} = 100 - 50 = 50\ V$$
So, the electric current passing through it,
$$i_{1} = \dfrac {50}{1000} = 50\ mA$$
So, current through Zener diode,
$$i_{2} = 50 - 25 = 25\ mA$$