In the circuit shown in Figure, if the diode forward voltage drop is 0.3 V the voltage difference between A and B is

At the middle right of the circuit the capacitor behaves like an open circuit for dc 0.2 mA current, so current will allow from A to  B only. Let potential across A and B is V, so by Kirchhoff's loop law
$\begin{array}{l}{\mathrm{V}}_{\mathrm{AB}}=\left(5,000\times 0.2\times {10}^{-3}\right)+0.3+5,000\times 0.2\times {10}^{-3}\\ \\ {\mathrm{V}}_{\mathrm{AB}}=1\mathrm{V}+0.3\mathrm{V}+1\mathrm{V}\\ {\mathrm{V}}_{\mathrm{AB}}=2.3\mathrm{V}\end{array}$