A resistor is loading a 6 V battery, drawn in the schematic as ideal. An ammeter is reading a current of 56mA and a voltmeter is purportedly reading a voltage of 5.92 V. No problem so far...
Part b: It is stated that the ammeter has a resistance of 10 ohm and the voltmeter has a 6 mA current draw. It turns out that this cannot really be the case!
56 mA through the 10 ohm of the ammeter means that it must drop 0.56 V; thus, the voltage measured by the voltmeter must be 5.44 V (not 5.92 V). 6 mA through the voltmeter means there is 50 mA through the unknown resistor; 5.44 V across it means the resistance must be 109 ohm. Contrast this to the answer of 118 ohm found from believing that there was 50 mA current through the resistor for a 5.92 V drop.
Part c:. How can the voltmeter possibly still be reading a voltage of 5.92 V when it is connected directly across the battery of 6 V? It can't happen unless the battery has an internal resistance! For a difference between the emf and the terminal voltage of 0.08 V, there must be an internal resistance of 1.29 ohm, given that the current flow of 62 mA now present (56 mA through the ammeter and 6 mA through the voltmeter).
Thanks to Celestina Laboy for calling this to my attention! [10/4/98]