Answer
$B=1.67 \times 10^{-20} \mathrm{~T} .$
Work Step by Step
The integral shown above will no longer (since now $r>R$ ) have $r$ as the upper limit; the upper limit is now $R$. Thus,
$$
\Phi_E=t \pi\left(\frac{1}{2} R^2-\frac{R^3}{3 R}\right)=\frac{1}{6} t \pi R^2 .
$$
Consequently, Eq. 32-3 becomes
$$
B(2 \pi r)=\frac{1}{6} \varepsilon_0 \mu_0 \pi R^2
$$
which for $r=0.0500 \mathrm{~m}$, yields
$$
B=\frac{\varepsilon_0 \mu_0 R^2}{12 r}=\frac{\left(8.85 \times 10^{-12}\right)\left(4 \pi \times 10^{-7}\right)(0.030)^2}{12(0.0500)}\\=1.67 \times 10^{-20} \mathrm{~T} .
$$
