A circuit contains a 3 V battery connected in series with a 10...

When the switch $S$ is closed, it creates a short circuit across the $20 \text{ } \Omega$ resistor.

Therefore, no current flows through the $20 \text{ } \Omega$ resistor, and the effective resistance of the circuit is only due to the $10 \text{ } \Omega$ resistor.

Effective resistance, $R_{eq} = 10 \text{ } \Omega$

Voltage of the battery, $V = 3 \text{ V}$

Using Ohm's law, the current $I$ through the circuit (and the $10 \text{ } \Omega$ resistor) is given by:

$I = \dfrac{V}{R_{eq}} = \dfrac{3}{10} = 0.3 \text{ A}$