The voltmeter shown in the circuit reads:

The voltmeter reads the voltage across the 250 MΩ resistors.

The resistors are connected in series. 

R_eq = R₁ + R₂

Req = (250 + 250) MΩ

Req = 500 MΩ

Current flowing in the circuit,

I = V/R = 24/500 = 0.048 μA

Voltage read by voltmeter will be:

0.048 μA × 250 MΩ = 12 V

Alternate Method:

By using voltage division rule:

The voltage across the resistor R₁ is:

\({V_{R1}} = V\left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)\)

The voltage across the resistor R₁ is:

\({V_{R2}} = V\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)

From the given figure,

R₁ = R₂ = 250 MΩ

V = 24 V

The voltage across the resistor R₂ is:

\({V_{R2}} = \frac{{24}}{{500}} \times 250 = 12\;{\rm{V}}\)