Answer
$B_{int}=19\;mT$
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, $hf=2\mu_z(B_{ext}+B_{int})$
or, $B_{int}=\frac{hf}{2\mu_z}-B_{ext}$
Substituting given values, we get
$B_{int}=\frac{6.63\times 10^{-34}\times34\times10^{6}}{2\times1.41\times 10^{-26}}-0.78$
or, $B_{int}=(0.799-0.78)\;T$
or, $B_{int}=0.019\;T$
or, $B_{int}=19\;mT$
$\therefore\;$ The magnitude of $B_{int}=19\;mT$
