Answer
$51\;mT$
Work Step by Step
The energy required of a photon to spin-flip the proton between the two orientations is
$hf=2\mu_zB$
or, $\frac{hc}{\lambda}=2\mu_zB$
or, $B=\frac{hc}{2\mu_z\lambda}$
For ground state hydrogen, $\mu_z=1\;\mu_B=9.27\times 10^{-24}\;J/T$
Substituting the gives values, we get,
$B=\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{2\times9.27\times 10^{-24}\times21\times 10^{-2}}\;T$
or, $B=0.051\;T$
or, $B=51\;mT$
$\therefore\;$ The effective magnitude of magnetic field as experienced by the electron in the ground-state hydrogen atom is $51\;mT$
