I know I'm a little late, but I'll take a shot at answering this for you. I'm actually very much a beginner at understanding electronics myself, so everyone: please keep me honest!
There has been some criticism of your question, as it does not show a complete circuit. I need to agree with this, as any reliable calculations within a circuit require knowledge of the entire circuit.
In order to know how much voltage will drop across a resistor, we need to know the value of the current flowing through said resistor. In order to know that, however, we need to go back even further and figure out the voltage being supplied to the circuit and the total resistance within the circuit. Once we have all our prerequisite information, we can say that the voltage drop across a resistor is equivalent to the charge's rate of flow (current) multiplied by the resistance it encounters. This holds true whether you are looking at a series circuit, a parallel circuit, or a mix of the two.
Take for example a circuit where a 12V battery serves as the power supply, and a 6 ohm resistor is placed in between the high and low potential terminals of the battery. The total current supplied by the battery will be equivalent to the voltage divided by the resistance (I = V / R = 12 / 6), resulting in a current of 2A. Now that we know the total resistance of the circuit and the value of the current, we can deduce that the voltage drop will be 12V (V = I * R = 2 * 6 = 12).
Now let's add a second resistor to the circuit (in series, right after the previous one), again with a resistance value of 6 ohms; the total resistance in the circuit is now 12 ohms. Since the resistance has changed, the amount of current will also change. As with our previous calculation, we will determine the total current by dividing the supplied voltage by the total resistance in the circuit. This time, our current is 1A (I = V / R = 12 / 12 = 1). Since the total current has changed, we need to recalculate our the amount of voltage consumed by each resistor. For each resistor (since they have identical resistance values), the total voltage drop is now 6V (V = I * R = 1 * 6 = 6). Since there are two resistors, the total voltage drop is 12V, which is equivalent to the supplied voltage.
This is an example of an ideal circuit, of course. In the real world, you'll see a lot more decimal places as the result of wires having their own internal resistance, resistors being accurate within certain tolerance levels, etc.