Answer:
hello your question is incomplete attached below is the complete and the required circuit diagrams 
answer :
Ai) <em>This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well</em>
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature 
hence At 0 mA current, there won't be any noticeable change
Explanation:
Ai) The voltage across the resistor will double when you double the current through the resistor
Given that : V = I*R.  
lets assume : I = 2 amperes , R = 3 ohms 
V = 2*3 = 6 v
secondly lets assume double the value of  (I)   i.e. I = 4 amperes 
hence : V = 4*3 = 12 volts 
<em>This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well</em>
Aii) Showing the two data points from simulation 
I1                    V1            I2= 2I1         V2=2V1       V1/ I1 =V2/I2
0.9*10^3     9 * 10^3     1.8*10^3       18*10^3          10 ohms
1.6 * 10^3    16 * 10^3    3.2*10^3     32*10^3         10 ohms
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature 
hence At 0 mA current, there won't be any noticeable change