Answer
See work below.
Work Step by Step
The manufacturer claims that 85 percent of the electrical power used by the heater is sent into the water as heat.
$$0.85\;IV=\frac{Q_{absorbed}}{t}=\frac{mc\Delta T}{t}$$
$$I=\frac{mc\Delta T}{0.85Vt}=\frac{(0.120kg)(4186J/kg)(70C)}{0.85(12V)(480s)}\approx 7.2A$$
b. Use Ohm’s law to find the resistance.
$$R=\frac{V}{I}=\frac{12V}{7.182A}=1.7\Omega$$