Correct answer is option (c) –1.51 eV
Explaination::
\(2 ^{nd}\ excited \ state \ n=3 E n =− \frac{ 13.06}{n^2} \) -{By formula }
\(E_3=-\frac{13.06}{3^2}=-\frac{13.06}{9}\)
= -1.51 eV