In the circuit R₁ = 400 ohm and R₂= 200 ohm. If the resistance of the voltmeter R =10,000 ohm, voltmeter reading is

Since, resistance of voltmeter  >>>  R₂, it can be neglected.

$\therefore R_1, R_2$ are in series.

Current in circuit will be 

$i = \frac{E}{R_1+R_2}=\frac{6}{200+400}=0.01A$

Voltage across R₂ is 

$V = i{R}_2=0.01\times 200=2V$

Due to the presence of voltmeter resistance parallel to R₂, there could be slight decrease in this voltage. 

So, $V<2v$