The correct option is A 121.6 nm
For Balmer series, 1λ=R(122−1n2)=R(14−1n2);n=3,4,5⋯
The second line in balmer series corresponds ot n= 4.
Hence, 1λ2=R(14−116)=3R16⇒λ2=163R
The wavelength of the first line (n = 2) in Lyman series is given by 1λ1=R(1−122)=R(1−14)=3R4⇒λ1=43R
∴λ1λ2=43R×3R16=14⇒λ1=λ24=486.44=121.6 nm