Two metre scales, one made of steel and the other of aluminium,...

Let $L_0$ be the true length of $1\text{ cm}$ on both scales at 20^{}^{{}^{\circ}} C.

At 0^{}^{{}^{\circ}} C, the temperature has decreased by \Delta T = 20^{}^{{}^{\circ}} C.

The true length of $1\text{ cm}$ on the aluminium scale at 0^{}^{{}^{\circ}} C is:
$L_{Al} = L_0(1 - \alpha_{Al}\Delta T)$

The true length of $1\text{ cm}$ on the steel scale at 0^{}^{{}^{\circ}} C is:
$L_{steel} = L_0(1 - \alpha_{steel}\Delta T)$

The ratio of aluminium-centimetre to steel-centimetre at 0^{}^{{}^{\circ}} C is:
$\text{Ratio} = \dfrac{L_{Al}}{L_{steel}} = \dfrac{1 - \alpha_{Al}\Delta T}{1 - \alpha_{steel}\Delta T}$

Using binomial approximation since $\alpha\Delta T \ll 1$:
$\text{Ratio} \approx (1 - \alpha_{Al}\Delta T)(1 + \alpha_{steel}\Delta T) \approx 1 - (\alpha_{Al} - \alpha_{steel})\Delta T$

Substituting the given values:
\alpha_{Al} = 2.3 \times 10^{-5}^{}^{{}^{\circ}} C^{-1}
\alpha_{steel} = 1.1 \times 10^{-5}^{}^{{}^{\circ}} C^{-1}
\Delta T = 20^{}^{{}^{\circ}} C

$\text{Ratio} = 1 - (2.3 \times 10^{-5} - 1.1 \times 10^{-5}) \times 20$

$\text{Ratio} = 1 - (1.2 \times 10^{-5}) \times 20$

$\text{Ratio} = 1 - 24 \times 10^{-5} = 1 - 0.00024 = 0.99976$