A potentiometer wire is 400 cm long and has a resistance of 8.0 Ω…

Step 1. The gradient. $I = \dfrac{3.0\ \text{V}}{8.0\ \Omega + 22\ \Omega} = 0.10\ \text{A}$. The wire has $0.10\ \text{A} \times 8.0\ \Omega = 0.80\ \text{V}$ across it, so $k = \dfrac{0.80\ \text{V}}{400\ \text{cm}} = 2.0\ \text{mV/cm}$.

Step 2. The $0.76\ \text{V}$ cell. $l = \dfrac{760\ \text{mV}}{2.0\ \text{mV/cm}} = 380\ \text{cm}$. That is on the wire.

Step 3. The new gradient. For $0.96\ \text{V}$ to balance at $320\ \text{cm}$, $k = \dfrac{960\ \text{mV}}{320\ \text{cm}} = 3.0\ \text{mV/cm}$. The wire needs $3.0\ \text{mV/cm} \times 400\ \text{cm} = 1.2\ \text{V}$ across it.

Step 4. The rheostat. The current is $\dfrac{1.2\ \text{V}}{8.0\ \Omega} = 0.15\ \text{A}$. The total resistance is $\dfrac{3.0\ \text{V}}{0.15\ \text{A}} = 20\ \Omega$. So the rheostat is $20\ \Omega - 8.0\ \Omega = 12\ \Omega$.