1. Lamp rating and required operating conditions

The arc lamp needs:

• Current: $I = 10\,\text{A}$
• Voltage across lamp: $V_R = 80\,\text{V}$

Since it works on direct current at $80\,\text{V}$ and $10\,\text{A}$, we treat the lamp as an effective resistance:

$R = \frac{V_R}{I} = \frac{80}{10} = 8\,\Omega$

2. AC supply and series inductor

The lamp is connected to:

$V = 220\,\text{V (rms)}, \quad f = 50\,\text{Hz}$

A pure inductor is connected in series so that the current remains $10\,\text{A}$.

For a series $R$-$L$ circuit, the impedance is:

$Z = \frac{V}{I} = \frac{220}{10} = 22\,\Omega$

Also,

$Z = \sqrt{R^2 + X_L^2}$

where $X_L = \omega L$.

So,

$22^2 = 8^2 + X_L^2$

$484 = 64 + X_L^2$

$X_L^2 = 420$

$X_L = \sqrt{420} \approx 20.49\,\Omega$

3. Find inductance

Angular frequency:

$\omega = 2\pi f = 2\pi \times 50 = 100\pi\,\text{rad/s}$

Thus,

$L = \frac{X_L}{\omega} = \frac{20.49}{100\pi}$

$L \approx \frac{20.49}{314} \approx 0.065\,\text{H}$

4. Match with options

The closest option is:

$\boxed{0.065\,\text{H}}$

So the correct option is B.