The energy of incident photon $$= (hc/ \lambda)$$
If $$W_i$$ is the ionization energy and $$E_k$$ the kinetic energy of emitted electron, then we have
$$\dfrac{hc}{\lambda_1} = W_1 + E_{K_1}$$
Therefore, for incident photons of wavelength $$\lambda_1 = 800 \mathring{A} = 8 \times 10^{-8} m$$
$$\dfrac{hc}{\lambda} = W_1 + E_{K_1}$$ ___(i)
And for incident photon of wavelength
$$\lambda_2 = 700 \mathring{A} = 7 \times 10^{-8} m$$,
$$\dfrac{hc}{\lambda} = W_1 + E_{K_2}$$ ____(ii)
Subtracting Eqs. (ii) from (i), we get
$$hc \left(\dfrac{1}{\lambda_2 } - \dfrac{1}{\lambda_1} \right) = E_{K_2} - E_{K_1}$$
or $$h = \dfrac{(E_{K_2} - E_{K_1}) \lambda_1 \lambda_2}{c(\lambda_1 - \lambda_2)}$$
Here, $$E_{K_1} = 1.8 eV = 1.8 eV = 1.8 \times 1.6 \times 10^{-19} J$$
and $$E_{K_2} = 4.0 eV = 4.0 \times 1.6 \times 10^{-19} J$$
Substituting given values, we get
$$h = \dfrac{(4.8 - 1.8) \times 1.6 \times 10^{-19} \times 8 \times 10^{-8} \times 7 \times 10^{-8}}{3 \times 10^{-8} (8 \times 10^{-8} - 7 \times 10^{-8})}$$
$$= 6.57 \times 10^{-34} Js$$