In a silicon diode, the reverse current increases from 10$\mathrm{\mu }$A to 20$\mathrm{\mu }$A , when the reverse voltage changes from 2V to 4V. The reverse ac resistance of the diode is 

Reverse bias often refers to a circuit's employment of a diode. When a diode is reverse biassed, the cathode's voltage is greater than the anode's. Therefore, no current will flow until the diode fails due to a strong enough electric field.
Drift current is the term used for this current.
There is a voltage drop from 4V to 2V.

Thus the change in voltage is, $△V=4V-2V$

Also change in current

 $$\begin{gathered} ∆I=20\mu A-10\mu A \\ ∆I=10\mu A \end{gathered}$$

We now that , $V=IR$

$R=2\times {10}^5\Omega$

Hence the correct answer is $2\times {10}^5\Omega .$