Answer
0.40.
Work Step by Step
The power delivered to the starter is $P=I^2R_S$.
$$\frac{P}{P_o}=\frac{I^2R_S}{I_o^2R_S}=(\frac{I}{I_o})^2$$
The current flowing is the battery emf divided by the equivalent resistance. The resistors are all in series, so the equivalent resistance is the sum of the resistances.
$$\frac{P}{P_o}= (\frac{\epsilon/R_{eq}}{\epsilon/R_{o\;eq}})^2= (\frac{R_{o\;eq}}{R_{eq}})^2$$
$$\frac{P}{P_o}= (\frac{0.02\Omega+0.15\Omega}{0.02\Omega+0.15\Omega +0.10\Omega})^2=0.40$$