Effects of using voltmeter w/ 20 kΩ impedance across 35 kΩ resistor?
Instead of claiming you can't do this, stop and actually think about it.

Draw the schematic of the circuit without the voltmeter.  Then draw the schematic with the voltmeter.  To the circuit, the voltmeter just looks like a 20 k&Omega; resistor.

Note that you are told that the 35 k&Omega; is in series with another 35 k&Omega; resistor.  Basically, the 35 k&Omega; resitor you are to measure the voltage on is being driven by a Thevenin source with 35 k&Omega; impedance.

You really should be able to solve this easily from here.  I expect the professor didn't get back to you because he thinks this is a easy problem you should be able to figure out yourself with a little thought.  He is right.