1. Understand the situation

A metre scale alone balances at its centre, i.e. at the $50.0\,\text{cm}$ mark. Hence its centre of mass is at $50.0\,\text{cm}$.

Two coins of mass $10\,\text{g}$ each are placed together at the $10.0\,\text{cm}$ mark.

So total mass of coins: $m_c = 20\,\text{g} = 0.02\,\text{kg}$

After placing the coins, the system balances at the $40.0\,\text{cm}$ mark. This means the combined centre of mass is at $40.0\,\text{cm}$.

Let the mass of the metre scale be $M$ kg.

2. Use centre of mass formula

Taking positions in cm:

• coins at $10$
• metre scale at $50$
• combined centre of mass at $40$

Hence, $\frac{M\cdot 50 + 0.02\cdot 10}{M+0.02} = 40$

3. Solve the equation

$50M + 0.2 = 40(M+0.02)$

$50M + 0.2 = 40M + 0.8$

$10M = 0.6$

$M = 0.06\,\text{kg}$

4. Match with the required form

Given mass of metre scale is $x \times 10^{-2}\,\text{kg}$

But $0.06\,\text{kg} = 6 \times 10^{-2}\,\text{kg}$

So, $x = 6$

5. Final answer

The required integer is: $\boxed{6}$