Answer
$$ \boldsymbol{B} = 0.5331~\text{T} $$
Work Step by Step
$\textbf{First}, \text{we calculate the energy of the photon using eq (38.2) as follows:}$
$$
\begin{align*}
E_{\gamma} &= hf\\
& = \left( 6.626\times 10^{-34}~\text{J}\cdot\text{s} \right) \times (22.7 \times 10^{6}~\text{s}^{-1})\\
& = 1.50\times 10^{-26}~\text{J}
\end{align*}
$$
$\text{The energy of the photon equals the energy difference between the two states ( in which$ \\
$ the nuclear spin component is parallel and antiparallel to the field ) }$
$$ \Delta E = E_\gamma = 1.50\times 10^{-26}~\text{J}$$
$\text{The energy of the state in which the spin component is parallel to the field is given by:} $
$$ U_{\parallel} = -|\mu_z| B$$
$\text{And, the energy of the state in which the spin component is anti-parallel to the field is given by:} $
$$ U_{\nparallel} = |\mu_z| B$$
$\text{Clearly, we can see that: }$
$$ U_{\parallel} = -U_{\nparallel}$$
$\text{Thus, :}$
$$
\begin{align*}
\Delta E &= U_{\nparallel} - U_{\parallel} = 2U_{\nparallel}\\
U_{\nparallel} &= \dfrac{\Delta E}{2}\\
& = \dfrac{1.50\times 10^{-26}~\text{J}}{2}\\
& = 7.52\times 10^{-27}~\text{J}
\end{align*}
$$
$\text{When the spin component of the proton is anti-parallel to the magnetic field, the interaction energy is given by:}$
$$ U_{\nparallel} = \mu_z B $$
$\text{Thus,}$
$$
\begin{align*}
B &= \dfrac{U_{\nparallel}}{\mu_z} = \dfrac{7.52\times 10^{-27}~\text{J}}{(2.7928)(5.05078\times 10^{-27}~\text{J/T})}\\\\
& = 0.5331~\text{T}
\end{align*}
$$
