Identify the circuit configuration.

The circuit consists of two main parallel branches connected to the 60V battery.

• The top branch contains a $3\Omega$ resistor in series with a parallel combination of $4\Omega$ and $12\Omega$ resistors.
• The bottom branch contains a $6\Omega$ resistor.

Let's call this $R_{p1}$.

$$R_{p1} = \frac{R_1 \times R_2}{R_1 + R_2}$$ $$R_{p1} = \frac{4\Omega \times 12\Omega}{4\Omega + 12\Omega} = \frac{48\Omega^2}{16\Omega} = 3\Omega$$

Let's call this $R_{top}$.
The $3\Omega$ resistor is in series with $R_{p1}$.

$$R_{top} = 3\Omega + R_{p1} = 3\Omega + 3\Omega = 6\Omega$$

Let's call this $I_{top}$.
The voltage across the top branch is the battery voltage, $V = 60\text{ V}$.

$$I_{top} = \frac{V}{R_{top}} = \frac{60\text{ V}}{6\Omega} = 10\text{ A}$$

Let's call this $V_{p1}$.
This voltage is the product of the current through the top branch and the equivalent resistance of the parallel part.

$$V_{p1} = I_{top} \times R_{p1} = 10\text{ A} \times 3\Omega = 30\text{ V}$$

This is the voltage across both the $4\Omega$ and the $12\Omega$ resistors.

Let's call this $P_{4\Omega}$.
We can use the formula $P = \frac{V^2}{R}$.

$$P_{4\Omega} = \frac{V_{p1}^2}{4\Omega} = \frac{(30\text{ V})^2}{4\Omega} = \frac{900\text{ V}^2}{4\Omega} = 225\text{ W}$$

Energy $E = P \times t$.

$$E_{4\Omega} = P_{4\Omega} \times t = 225\text{ W} \times 2\text{ s} = 450\text{ J}$$

The energy dissipated by the $4\Omega$ resistor in 2 seconds is $450\text{ J}$.

The correct option is B.
$\boxed{\text{B}}$