1. Use colligative property relations

For the same solution, $\Delta T_f = K_f m$ and $\Delta T_b = K_b m$ where $m$ is the molality.

2. Given freezing point depression

The freezing point of the solution is $-0.186^\circ\text{C}$.

Since pure water freezes at $0^\circ\text{C}$, the depression in freezing point is $\Delta T_f = 0 - (-0.186) = 0.186^\circ\text{C}$

3. Find molality relation

From $\Delta T_f = K_f m$ we get $m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86}$

4. Calculate boiling point elevation

Now, $\Delta T_b = K_b m = K_b\left(\frac{\Delta T_f}{K_f}\right)$

So, $\Delta T_b = 0.512 \times \frac{0.186}{1.86}$

Since $\frac{0.186}{1.86} = 0.1$ therefore, $\Delta T_b = 0.512 \times 0.1 = 0.0512^\circ\text{C}$

5. Match with options

$\boxed{\Delta T_b = 0.0512^\circ\text{C}}$

So the correct option is B.