From Einstein's photoelectric equation,

$$\begin{gathered} e{V}_s=h\nu -\varphi \\ \Rightarrow V_s=\left(\frac{h}{e}\right)\nu -\frac{\varphi }{e} \end{gathered}$$

Clearly, it is a linear relation between stopping potential and frequency of the incident light.

Therefore, slope of the curve is $=\frac{h}{e}$.

From question,

$$\begin{gathered} \Rightarrow \frac{h}{e}=4\times {10}^{-15} \\ \Rightarrow h=\left(4\times {10}^{-15}\right)\times \left(1.6\times {10}^{-19}\right) \\ =6.4\times {10}^{-34} \mathrm{J} \mathrm{s} \end{gathered}$$