For visible line spectrum,(i.e) Balmer series $n_1=2$.Also for minimum energy transition $n_2=3$
$\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{2^2}-\frac{1}{3^2}\bigg]$
$\Rightarrow 1.1\times 10^7\big[\large\frac{1}{4}-\frac{1}{9}\big]$
$\Rightarrow 1.1\times 10^7\times \large\frac{5}{36}$
$\lambda=6.55\times 10^{-7}m$
Now $E=\large\frac{hc}{\lambda}$
$\Rightarrow \large\frac{6.62\times 10^{-34}\times 3\times 10^8}{6.55\times 10^{-7}}$
$\Rightarrow 3.03\times 10^{-19}J$
If N electrons show this transition in 1g atom of H then
Energy released=$E\times N$
$\Rightarrow 3.03\times 10^{-19}\times 6.023\times 10^{23}$
$\Rightarrow 18.25\times 10^4$
$\Rightarrow 182.5KJ$
Hence (a) is the correct answer.