Two metals A and B give parallel lines on a graph of stopping potential against frequency…

Step 1. Metal A. $V_0 = 4.14 \times 10^{-15}\ \text{V s} \times (8.0 - 5.0) \times 10^{14}\ \text{Hz} = 1.24\ \text{V}$.

Step 2. The work functions. $\phi_A = 4.14 \times 10^{-15} \times 5.0 \times 10^{14} = 2.07$ eV. So $\phi_B = 2.07 + 1.2 = 3.27$ eV.

Step 3. Metal B. The photon energy is $4.14 \times 10^{-15} \times 8.0 \times 10^{14} = 3.31$ eV. So $V_0 = 3.31 - 3.27 = 0.042$ V.

Step 4. The matching frequency. B needs $1.2$ eV more photon energy than A. The extra frequency is $\dfrac{1.2\ \text{eV}}{4.14 \times 10^{-15}\ \text{eV s}} = 2.9 \times 10^{14}$ Hz. So $\nu = 8.0 \times 10^{14} + 2.9 \times 10^{14} = 10.9 \times 10^{14}$ Hz.