1. Use the power dissipation to find the resistance of the coil

The coil is an inductor with some resistance, so the given dissipation is due to its resistance:

$P = I^2 R$

Given:

• $P = 640\,\text{W}$
• $I = 8\,\text{A}$

So,

$R = \frac{P}{I^2} = \frac{640}{8^2} = \frac{640}{64} = 10\,\Omega$

2. Use stored magnetic energy to find the inductance

Energy stored in an inductor is:

$U = \frac{1}{2} L I^2$

Given:

• $U = 64\,\text{J}$
• $I = 8\,\text{A}$

Thus,

$64 = \frac{1}{2} L (8^2) = \frac{1}{2} L (64) = 32L$

$L = \frac{64}{32} = 2\,\text{H}$

3. Find the time constant of the RL circuit

For a coil connected across an ideal battery, the RL time constant is:

$\tau = \frac{L}{R}$

Substitute $L = 2\,\text{H}$ and $R = 10\,\Omega$:

$\tau = \frac{2}{10} = 0.2\,\text{s}$

4. Match with the options

$\boxed{0.2\,\text{s}}$

So the correct option is D.