A 9.0 Ω resistor X is in the left gap of a meter bridge and a 6.0 Ω box is in the right gap…

Step 1. Before. $l = \dfrac{100X}{X + R} = \dfrac{100 \times 9.0}{9.0 + 6.0} = 60.0\ \text{cm}$.

Step 2. The new right arm. $\dfrac{6.0 \times 12}{6.0 + 12}\ \Omega = 4.0\ \Omega$.

Step 3. After. $l = \dfrac{100 \times 9.0}{9.0 + 4.0} = 69.2\ \text{cm}$.