C

Step 1: Given that the voltmeter reads 1.52 V when the switch S is open, this reading represents the emf \( \varepsilon \) of the battery. Hence, \( \varepsilon = 1.52 \text{ V} \).

Step 2: When the switch is closed, the voltmeter reads 1.45 V which shows the terminal voltage \( V \) across the battery while providing current. The ammeter reads 1.0 A, which is the current \( I \).

Step 3: The relationship between emf, terminal voltage, internal resistance, and current is given by the formula: \( V = \varepsilon - Ir \).

Step 4: Rearranging the formula for \( r \):
\( r = \frac{\varepsilon - V}{I} \)
Substitute the values into the equation.
\( \varepsilon = 1.52 \text{ V} \), \( V = 1.45 \text{ V} \), and \( I = 1.0 \text{ A} \):
\( r = \frac{1.52 - 1.45}{1.0} \)
\( r = \frac{0.07}{1.0} \)
\( r = 0.07 \text{ ohms} = 70 \text{ m}\Omega \)

Therefore, the internal resistance of the battery is 70 mΩ.