What does it mean by 'photons of higher intensity'?
The assertion is incorrect. Intensity is related to the number of photons emitted and has nothing to do with Bohr orbit transitions.  

The reason is correct. Rather than higher intensity, photons of higher energy are emitted when an electron falls from the 2nd to the 1st Bohr orbit as compared to when it falls from the 4th to the 2nd orbit.  This is because the energy emitted or absorbed in a Bohr transition is given by the following equation where Z is the atomic number of the atom, R is a constant and n represents the quantum number of the final and initial Bohr orbits.

$\mathrm{\Delta E=-Z^2R \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right )}$

since

$\mathrm{\frac{1}{1^2}-\frac{1}{2^2}=3/4}$ 

is greater than 

$\mathrm{\frac{1}{2^2}-\frac{1}{4^2}=3/16}$ 

more energy is emitted in the first case than the second.