Calculate the moles of hydrochloric acid (HCl) used.

Volume of HCl $= 200 \text{ cm}^3 = 0.200 \text{ L}$
Concentration of HCl $= 1 \text{ M} = 1 \text{ mol/L}$

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 1 \text{ mol/L} \times 0.200 \text{ L} = 0.200 \text{ mol}$$

The balanced equation is: $\text{BCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{BCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$
From the equation, 1 mole of $\text{BCO}_3$ reacts with 2 moles of HCl.

$$\text{Moles of BCO}_3 = \frac{\text{Moles of HCl}}{2} = \frac{0.200 \text{ mol}}{2} = 0.100 \text{ mol}$$

Mass of $\text{BCO}_3 = 1 \text{ g}$

$$\text{Molar mass of BCO}_3 = \frac{\text{Mass}}{\text{Moles}} = \frac{1 \text{ g}}{0.100 \text{ mol}} = 10.0 \text{ g/mol}$$

The molar mass of $\text{BCO}_3$ is the sum of the relative atomic masses of B, C, and three O atoms.
Let the relative atomic mass of B be $x$.

$$x + \text{C} + (3 \times \text{O}) = 10.0$$

Given C = 12.0 and O = 16.0:

$$x + 12.0 + (3 \times 16.0) = 10.0$$ $$x + 12.0 + 48.0 = 10.0$$ $$x + 60.0 = 10.0$$

This result indicates an error in the problem statement or given values, as the calculated molar mass of $\text{BCO}_3$ (10.0 g/mol) is less than the mass of the carbonate part ($\text{CO}_3$) alone (12.0 + 3*16.0 = 60.0 g/mol). This would imply a negative atomic mass for B, which is not possible.

Let's re-evaluate the calculation assuming the molar mass of $\text{BCO}_3$ should be greater than 60.0 g/mol.
If the mass of the carbonate was, for example, 10 g instead of 1 g, then:
Molar mass of $\text{BCO}_3 = \frac{10 \text{ g}}{0.100 \text{ mol}} = 100 \text{ g/mol}$
Then, $x + 60.0 = 100.0$
$x = 100.0 - 60.0 = 40.0$

Given the problem as stated, with 1 g of carbonate reacting, the calculated molar mass of $\text{BCO}_3$ is 10.0 g/mol. This value is chemically impossible for a compound containing C and 3 O atoms. There might be a typo in the mass of the carbonate (1 g) or the concentration/volume of HCl.

Assuming the question intends for a positive atomic mass for B, and that the molar mass of $\text{BCO}_3$ should be greater than 60.0 g/mol, there is an inconsistency in the provided data. If we proceed strictly with the given numbers, the result for B would be negative.

Let's assume the molar mass of $\text{BCO}_3$ was intended to be a value that yields a positive atomic mass for B. For example, if the mass of $\text{BCO}_3$ was 7 g instead of 1 g:
Moles of $\text{BCO}_3 = 0.100 \text{ mol}$
Mass of $\text{BCO}_3 = 7 \text{ g}$
Molar mass of $\text{BCO}_3 = \frac{7 \text{ g}}{0.100 \text{ mol}} = 70 \text{ g/mol}$
Then, $x + 12.0 + (3 \times 16.0) = 70.0$
$x + 12.0 + 48.0 = 70.0$
$x + 60.0 = 70.0$
$x = 70.0 - 60.0 = 10.0$

Given the direct numbers in the problem, the calculation leads to an impossible result. If we strictly follow the numbers:

$$x = 10.0 - 60.0 = -50.0$$

Since atomic mass cannot be negative, there is an error in the question's numerical values. However, if we assume a common element like Calcium (Ca) for B, its atomic mass is 40.0. If B is Ca, then $\text{CaCO}_3$ has a molar mass of $40.0 + 12.0 + (3 \times 16.0) = 100.0 \text{ g/mol}$.
If the molar mass of $\text{BCO}_3$ is 100.0 g/mol, then 0.100 mol would be 10.0 g. This suggests the "1 g" might be a typo for "10 g".

Assuming the mass of the carbonate was intended to be 10 g (a common value for such problems to yield a reasonable answer):
Step 3 (Revised): Calculate the molar mass of $\text{BCO}_3$.
Mass of $\text{BCO}_3 = 10 \text{ g}$ (assuming typo from 1 g)

$$\text{Molar mass of BCO}_3 = \frac{10 \text{ g}}{0.100 \text{ mol}} = 100.0 \text{ g/mol}$$

Step 4 (Revised): Calculate the relative atomic mass of B.

$$x + 12.0 + (3 \times 16.0) = 100.0$$ $$x + 12.0 + 48.0 = 100.0$$ $$x + 60.0 = 100.0$$ $$x = 100.0 - 60.0 = 40.0$$

The relative atomic mass of B is 40.0.