In the given circuit, D_1 and D_2 are ideal p-n junction diodes....

In the given circuit, the positive terminal of the $12\text{ V}$ battery is connected to the left side.

Diode $D_1$ is forward-biased as its p-side faces the higher potential. Diode $D_2$ is reverse-biased as its n-side faces the higher potential.

Since $D_2$ is an ideal diode in reverse bias, it acts as an open circuit. The ideal voltmeter also has infinite resistance, so no current flows through the circuit ($I = 0$).

The voltage drop across the resistor $R$ is $V_R = I R = 0\text{ V}$.

The voltage drop across the ideal forward-biased diode $D_1$ is $0\text{ V}$.

Applying Kirchhoff's Voltage Law:

$12 - V_R - V_{D1} - V_{D2} = 0$

$12 - 0 - 0 - V_{D2} = 0$

$V_{D2} = 12\text{ V}$

The voltmeter reads the potential difference across $D_2$, which is $12\text{ V}$.

Answer: $12\text{ V}$