Explanation

(a) $W = MB(\cos\theta_{1} - \cos\theta_{2})$
When the magnet is rotated from $0{^\circ}$ to $60{^\circ}$, then work done is $0.8\ J$
$0.8 = MB\left( \cos{0{^\circ}} - \cos{60{^\circ}} \right) = \frac{MB}{2}$
$\Rightarrow MB = 1.6\ N - m$
In order to rotate the magnet through an angle of $30{^\circ},\ i.e.,$ from $60{^\circ}$ to $90{^\circ}$, the work done is
$W^{'} = MB\left( \cos{60{^\circ}} - \cos{90{^\circ}} \right) = MB\left( \frac{1}{2} - 0 \right)$
$= \frac{MB}{2} = \frac{1.6}{2} = 0.8J = 0.8 \times 10^{7}\ erg$