Wavelength of light at height y 
$\lambda =[4000+(8000-4000\left)\mathrm{y}\right]\text{Å}=4\times {10}^{-7}(1+\mathrm{y})$   meter.
Consider a strip of thickness dy at height y 
Energy incident on element per sec = Intensity x strip area = $\left(I\right) \left(\mathcal{l}dy\right)$
No. of photons incident per second on element  
$\mathrm{dn}=\frac{\mathrm{I}\ell \mathrm{dy}}{\frac{\mathrm{hc}}{\lambda }}=\frac{\mathrm{I}\ell \mathrm{dy}}{\mathrm{hc}}\left(4\times {10}^{-7}\right)(1+\mathrm{y})$ 
Total no. of electrons ejected due to photons per sec 
 $\mathrm{n}=\frac{\mathrm{I}\ell }{\mathrm{hc}}\left(4\times {10}^{-7}\right){\int }_0^{\mathcal{l}/2} (1+\mathrm{y})\mathrm{dy}=\frac{1\left(1\right)\left(4\times {10}^{-7}\right)}{\left(6.6\times {10}^{-34}\right)\left(3\times {10}^8\right)}\frac{5}{8}=\frac{20}{19.8}\times \frac{{10}^{19}}{8}$
$\approx \frac{{10}^{19}}{8}$ 
So k = 8
[Note: No electrons are ejected from upper half, because wavelength of incident wave becomes  greater than threshold wavelength]