High voltage!

Using Millman's Theorem:

${ V }_{ t }=\frac { \frac { 10 }{ 1 } +\frac { 4 }{ 2 } }{ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } } =\boxed { 8V }$

first of all we should know that no current will flow through the voltmeter because of the Rm(resistance multiplier) so the total voltage is 10-4= 6 V and the total current is 6/2+1 = 2 A so the potential difference across 2 ohm resistor is 4 V ∵ the 10 V battery is greater by 6 V ∴ potential diff. is 2 V and the voltmeter's reading is 10-2 = 8 V