Answer
The internal resistance of battery 1 is $~~0.033~\Omega$
Work Step by Step
We can find the current when $R = 0.10~\Omega$:
$i = \frac{\mathscr{E}}{R}$
$i = \frac{1.20~V+1.20~V}{0.10~\Omega}$
$i = 24~A$
When $R=0.10~\Omega$, note on the graph that $V = 0.40~V$ for battery 1
We can find the internal resistance:
$V = \mathscr{E}-ir$
$ir = \mathscr{E}-V$
$r = \frac{\mathscr{E}-V}{i}$
$r = \frac{1.20~V-0.40~V}{24~A}$
$r = 0.033~\Omega$
The internal resistance of battery 1 is $~~0.033~\Omega$
