Answer
$\left|\Phi_C\right|=47.4 \mu \mathrm{Wb}$.
Work Step by Step
Substituting the values given, the flux through the second end is
$
\Phi_2=\pi(0.120 \mathrm{~m})^2\left(1.60 \times 10^{-3} \mathrm{~T}\right)=+7.24 \times 10^{-5} \mathrm{~Wb}=+72.4 \mu \mathrm{Wb} .
$
Since the three fluxes must sum to zero,
$
\Phi_C=-\Phi_1-\Phi_2=25.0 \mu \mathrm{Wb}-72.4 \mu \mathrm{Wb}=-47.4 \mu \mathrm{Wb} .
\\
$
Thus, the magnitude is $\left|\Phi_C\right|=47.4 \mu \mathrm{Wb}$.
