The stopping voltage is 0.45 V for light of 6.0 × 10¹⁴ Hz, and 1.68 V for 9.0 × 10¹⁴ Hz…

Step 1. (a) The slope. $h = \dfrac{1.68\ \text{V} - 0.45\ \text{V}}{9.0 \times 10^{14}\ \text{Hz} - 6.0 \times 10^{14}\ \text{Hz}} = 4.1 \times 10^{-15}\ \text{eV s}$, which is $4.1 \times 10^{-15} \times 1.6 \times 10^{-19}\ \text{J s} = 6.56 \times 10^{-34}\ \text{J s}$.

Step 2. (b) $\phi = 4.1 \times 10^{-15}\ \text{eV s} \times 6.0 \times 10^{14}\ \text{Hz} - 0.45\ \text{eV} = 2.46\ \text{eV} - 0.45\ \text{eV} = 2.01\ \text{eV}$.