Answer
$ {\bf 2.10}\;\rm T$
Work Step by Step
We know that a cyclotron motion has a radius that is given by
$$r=\dfrac{mv}{qB}=\dfrac{mv}{eB}$$
Hence, the magnetic field is given by
$$B=\dfrac{mv}{er} $$
where $v=2\pi r/T$ where $T$ is the periodic time for one full cycle, which is given by $T=1/f_{\rm cyc}$, so
$$B=\dfrac{2\pi r f_{\rm cyc}m }{er}$$
$$B_1=\dfrac{2\pi f_{\rm cyc}m }{e } \tag 1$$
We know that Hall voltage is given by
$$(\Delta V_{\rm H})_1=\dfrac{I_1B_1}{tne}\tag 2$$
Hence,
$$tne =\dfrac{I_1B_1}{(\Delta V_{\rm H})_1}$$
Plug $B$ from (1),
$$tne =\dfrac{I_1 }{(\Delta V_{\rm H})_1}\dfrac{2\pi f_{\rm cyc}m }{e }\tag 3$$
Solving (2) for $B$ to find the new magnetic field.
$$B_2=\dfrac{(\Delta V_{\rm H})_2 tne}{I_2}$$
Plug from $tne$ (3),
$$B_2=\dfrac{(\Delta V_{\rm H} )_2 }{I_2}\dfrac{I_1 }{(\Delta V_{\rm H})_1}\dfrac{2\pi f_{\rm cyc}m }{e }$$
where $I_1=I_2$,
$$B_2= \dfrac{(\Delta V_{\rm H} )_2}{(\Delta V_{\rm H})_1}\dfrac{2\pi f_{\rm cyc}m }{e }$$
Plug the known;
$$B_2=\dfrac{(1.735)}{(0.543)} \dfrac{2\pi (10\times 10^6)(1.67\times 10^{-27}) }{(1.6\times 10^{-19}) }$$
$$B_2=\color{red}{\bf 2.10}\;\rm T$$
