Answer
See the detailed answer below.
Work Step by Step
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
Hence, the current needed is given by
$$I=\dfrac{2\pi d B}{\mu_0}$$
At point $d=0.01\;\rm m$;
$$I=\dfrac{2\pi (0.01) B}{4\pi \times 10^{-7}}$$
$$I=\dfrac{ B}{200 \times 10^{-7}}$$
Using the data in Table 32.1, to find the currents needed,
$$I_1=\dfrac{ B_{\rm Earth's \;surface}}{200 \times 10^{-7}}=\dfrac{ (5\times 10^{-5})}{200 \times 10^{-7}}=\color{red}{\bf 2.5}\;\rm A$$
$$I_2=\dfrac{ B_{\rm Refrigerator\;magnet}}{200 \times 10^{-7}}=\dfrac{ (5\times 10^{-3})}{200 \times 10^{-7}}=\color{red}{\bf 250}\;\rm A$$
$$(I_3)_{\rm minimun}=\dfrac{ B_{\rm Lab\;magnet}}{200 \times 10^{-7}}=\dfrac{ (0.1)}{200 \times 10^{-7}}=\color{red}{\bf 5000}\;\rm A$$
$$(I_3)_{\rm maximum}=\dfrac{ B_{\rm Lab\;magnet}}{200 \times 10^{-7}}=\dfrac{ (1)}{200 \times 10^{-7}}=\color{red}{\bf 50,000}\;\rm A$$
$$I_4=\dfrac{ B_{\rm superconducting \;magnet}}{200 \times 10^{-7}}=\dfrac{ (10)}{200 \times 10^{-7}}=\color{red}{\bf 500,000}\;\rm A$$
