1. Find the volume of ice

The pipe is of length $1\,\text{m}$ and cross-sectional area $A = 1\,\text{cm}^2 = 10^{-4}\,\text{m}^2$

So volume of ice is $V = A\ell = 10^{-4} \times 1 = 10^{-4}\,\text{m}^3$

2. Find the mass of ice

Given density of ice, $\rho = 10^3\,\text{kg m}^{-3}$

Hence, $m = \rho V = 10^3 \times 10^{-4} = 10^{-1}\,\text{kg} = 0.1\,\text{kg}$

3. Heat needed to raise temperature from $-10^\circ C$ to $0^\circ C$

Using $Q_1 = mc\Delta T$ with $c = 2\times 10^3\,\text{J kg}^{-1}\,{}^\circ\text{C}^{-1}, \quad \Delta T = 10^\circ C$

So, $Q_1 = 0.1 \times 2\times 10^3 \times 10 = 2000\,\text{J}$

4. Heat needed to melt the ice at $0^\circ C$

Using $Q_2 = mL$ where $L = 3.33\times 10^5\,\text{J kg}^{-1}$

Thus, $Q_2 = 0.1 \times 3.33\times 10^5 = 3.33\times 10^4\,\text{J} = 33300\,\text{J}$

5. Total heat required

$Q = Q_1 + Q_2 = 2000 + 33300 = 35300\,\text{J}$

6. Power produced by the resistor

Given current $I = 0.5\,\text{A}$ and resistance $R = 4\,\text{k}\Omega = 4000\,\Omega$

Power is $P = I^2R = (0.5)^2 \times 4000 = 0.25 \times 4000 = 1000\,\text{W}$

7. Time required

$t = \frac{Q}{P} = \frac{35300}{1000} = 35.3\,\text{s}$

8. Match with options

$\boxed{35.3\,\text{s}}$

So the correct option is B.