Answer
$r = 50~m \Omega$
Work Step by Step
In part (a), we found that the potential difference across the resistor is $1.0~V$
We can find the current in the circuit:
$i = \frac{V}{R}$
$i = \frac{1.0~V}{0.10~\Omega}$
$i = 10~A$
We can find the internal resistance of the battery:
$V = \mathscr{E}-i~r$
$i~r = \mathscr{E}-V$
$r = \frac{\mathscr{E}-V}{i}$
$r = \frac{1.5~V-1.0~V}{10~A}$
$r = 50~m \Omega$
