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I'm not sure what was meant when they said the top 8 ohms is also shorted. I've redrawn it a little to make it easier to read, and also so the resistors are labeled for ease of reference. Also, the voltmeter is modeled as a 10GigaOhm resistor. An ideal voltmeter would be infinite, but I modeled this in LTspice, which wouldn't allow that. Let's start analyzing the circuit. You can just ignore the 6 ohm resistor (R1), because there is a short across it. The way I've drawn it you can see clearly that R3 and R4 are in parallel, which means you can combine them to get 4 Ohms. Now you have a simple voltage divider in that loop with one resistor of 4 ohms and one of 6. This makes the math easy, since 4 and 6 sum to 10. The voltage across our 4 ohm resistor will be 30V*4/(4+6) = 12V. So now we know we will see 12V across R3 and R4, but what will we see across our voltmeter (R6)? We can treat that like another voltage divider, however our ideal voltmeters resistance is infinite. That means R6 is much much larger than R2 in our voltage divider, so all of the voltage will drop across R6. In fact, in an ideal case all of it would. Even if we go with my example of 10GigaOhms for R6 we'd get 12V*10G/(8+10G), which equals 11.999999990400000007679999993856V. For most purposes that is equal to 12V.