Explanation

(d) Number of spectral lines obtained due to transition of electrons from $n$th orbit to lower orbit is,
$\ N =$ $\frac{n(n - 1)}{2}$
I case $6\ =$ $\frac{n_{1}(n_{1} - 1)}{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{1} = 4$
II case $3\ =$ $\frac{n_{2}(n_{2} - 1)}{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{2} = 3$
Velocity of electron in hydrogen atom in $n$th orbit
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{n} \propto \frac{1}{n}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{n}}{{v^{'}}_{n}} = \frac{n_{2}}{n_{1}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n_{6}}{n_{3}} = \ \frac{3}{4}$