Several equipotential surfaces of a magnetic scalar potential are...

The relationship between the magnetic field $B$ and the magnetic scalar potential $V$ is given by $B = \dfrac{\Delta V}{\Delta r_{\perp}}$, where $\Delta r_{\perp}$ is the perpendicular distance between the equipotential surfaces.

From the given figure, the distance between two adjacent equipotential lines along the x-axis is:
$\Delta x = 20 \text{ cm} - 10 \text{ cm} = 10 \text{ cm} = 0.1 \text{ m}$

The equipotential lines are inclined at an angle of $30^{{}^{{}^{\circ}}}$ with the x-axis. The perpendicular distance $\Delta r_{\perp}$ between the adjacent equipotential lines is:
$\Delta r_{\perp} = \Delta x \sin(30^{{}^{{}^{\circ}}}) = 0.1 \times \dfrac{1}{2} = 0.05 \text{ m}$

The difference in magnetic scalar potential between these adjacent lines is:
$\Delta V = 0.2 \times 10^{-4} \text{ T m} - 0.1 \times 10^{-4} \text{ T m} = 0.1 \times 10^{-4} \text{ T m}$

The magnitude of the magnetic field is:
$B = \dfrac{\Delta V}{\Delta r_{\perp}} = \dfrac{0.1 \times 10^{-4}}{0.05} = 2.0 \times 10^{-4} \text{ T}$