1. Use the temperature dependence of resistance

For a conductor, $R_t = R_0(1+\alpha t)$ where $R_0$ is the resistance at $0^\circ\text{C}$.

Since the conductor is connected across the same source, voltage $V$ is constant, so $I_t = \frac{V}{R_t} = \frac{V}{R_0(1+\alpha t)}$ Thus, $I_t \propto \frac{1}{1+\alpha t}$

2. Given currents at two temperatures

At $0^\circ\text{C}$: $I_0 = 2\text{ A}$

At $100^\circ\text{C}$: $I_{100} = 1.2\text{ A}$

Now, $\frac{I_0}{I_{100}} = \frac{1+100\alpha}{1}$ So, $\frac{2}{1.2} = 1+100\alpha$ $\frac{5}{3} = 1+100\alpha$ $100\alpha = \frac{2}{3}$ $\alpha = \frac{1}{150}$

3. Find current at $50^\circ\text{C}$

$I_{50} = \frac{I_0}{1+50\alpha}$ Substitute $I_0=2$ A and $\alpha=\frac{1}{150}$: $I_{50} = \frac{2}{1+50\cdot \frac{1}{150}}$ $I_{50} = \frac{2}{1+\frac{1}{3}}$ $I_{50} = \frac{2}{\frac{4}{3}} = \frac{3}{2} = 1.5\text{ A}$

4. Convert to the required form

$1.5\text{ A} = 1500\text{ mA} = 15 \times 10^2\text{ mA}$

So the required integer is $\boxed{15}$