well the Balmer series is:
let w= wavelength
n = prinicpal quantum number SQUARED, thus n>4 & n = 9, 16, 25, etc
1/w = R (1/4-1/n) = R((n-4)/4n)
since balmer series ends on n=2
thus w=(1/R)*((4n)/(n-4))
since 1/R is constant, w will be largest (longest) when 4n/(n-4) is largest, which occurs for the smallest possible value of n (by substiuting values), the smallest n = 9, thus the 3rd orbital shell
thus sub in and u get :
w=(1/R)*(36/5)
= 36/5R
R=1.097*10^7
w=6.56*10^-7m
yes, i know it can be done without the formal mathematics, but its good practice!!!