The frequency of the nth-harmonic of a closed pipe resonator is given by
[tex]f_n = n f_1[/tex]
where [tex]f_1 [/tex] is the fundamental frequency.
For the pipe in our problem, the fundamental frequency is [tex]f_1 = 256 Hz[/tex], so the 4th harmonic (n=4) will be
[tex]f_4 = 4 \cdot 256 Hz = 1024 Hz[/tex]
