If the input is a simple dc source and it has been on for a "long time", then you are in dc steady state; therefore, the capacitor may be replaced by and open circuit. The voltage measured by U1 can be written down by inspection using voltage division
\[
U_1 = \frac{R_x||R_2}{R_1 + (R_x||R_2)} V_1
\]
where \[R_x||R_2\] is the parallel combination of \[R_x\] and \[R_2\]. For the values given \[R_x||R_2 \approx 9.09 \Omega\]. The numerical value of the voltage measured by the voltmeter is
\[
U_1 = \frac{9.09 \Omega}{1\text{k}\Omega + 9.09 \Omega} 12 \text{V} \approx 108.1 \text{mV}
\]
And it looks like this is exactly what is shown in your picture (0.108).
Best regards,
v_c