The peak current drawn by a DC motor with a commutator is determined by the resistance of the armature winding. When the motor is first switched on, it will draw:
$$I = Vs/Ra$$
where \$Vs\$ is the supply voltage and \$Ra\$ is the resistance of the armature winding.
The current with the shaft turning is:
$$I = (Vs - Vb)/Ra$$
where \$Vb\$ is the back EMF generated by the speed of the armature.
$$Vb = Speed \cdot K$$
where \$K\$ is the volt per RPM rating of the motor.
The heat developed in the armature winding is \$I^2 \cdot Ra\$ so the higher \$Ra\$, the more internal heat is produced in the motor. That means an efficient motor needs to have a low \$Ra\$ value and a high initial starting current.
If the motor is 90% efficient, the initial current will be 10 times the normal running current. While it is impossible to determine the maximum starting current without knowing the design details, it could easily be 3X rated current and is likely to be much more than that.
The peak starting current is an instantaneous value. As the motor speed increases from standstill, the current drops quickly.
The above analysis neglects the internal resistance of the battery. That is likely to be important. However the more the current is limited by the battery, the longer it will take the motor to accelerate. It may be difficult to determine if a wrong choice of battery will prevent the engine from starting or harm the battery.