Answer
The internal resistance of the battery is approximately 1.2Ω
Work Step by Step
Current across the resistor is,
$V=IR$
$8.5V=I\times 20Ω$
$I = 0.425A$
Same current applies across the internal resistance
Potential drop across the internal resistance can be taken as,
$9.0V-8.5V = 0.5V$
Hence the internal resistance can be found,
$0.5V = 0.425A \times R$
$R = 1.176Ω \approx 1.2Ω$
