Answer
$512\;\rm Hz$
Work Step by Step
We know, in the open-open tube, that the frequency is given by
$$f_m=\dfrac{mv}{2L}$$
and for the fundamental frequency,
$$f_1=\dfrac{ v}{2L}$$
For the helium case,
$$f_1=\dfrac{ v_{He}}{2L}$$
Hence,
$$2L=\dfrac{ v_{He}}{f_1}\tag 1$$
in the air case,
$$f_2=\dfrac{ v_{air}}{2L}$$
Plugging $2L$ from (1);
$$f_2=\dfrac{ v_{air}f_1}{v_{He}}$$
Plugging the known;
$$f_2=\dfrac{ (331)(1500)}{(970)}$$
$$f_2=\color{red}{\bf512}\;\rm Hz$$
