In a meter bridge the unknown resistance X is in the left gap and a 12 Ω box is in the right gap…

Step 1. The resistance. $X = R\,\dfrac{l}{100 - l} = 12\ \Omega \times \dfrac{57.5}{42.5} = 16.2\ \Omega$.

Step 2. After interchanging. Now $\dfrac{R}{X} = \dfrac{l'}{100 - l'}$, so $l' = \dfrac{100R}{R + X} = \dfrac{100 \times 12}{12 + 16.2} = 42.5\ \text{cm}$.