Solution:
The magnetic flux through the surface is given by \(\phi = B A \sin\theta\). Substituting \(\phi = 10^{-3}\text{ Wb}\), \(A = 0.02\text{ m}^2\), and \(\theta = 60^\circ\), we get \(B = \frac{10^{-3}}{0.02 \times \sin 60^\circ} \approx 0.058\text{ T}\).
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