Magnetic moment of the bar magnet, $M=5.25 \times 10^{-2} \,J\,T ^{-1}$

Magnitude of earth's magnetic field at a place, $H=0.42 \,G=0.42 \times 10^{-4} \,T$

$(a)$ The magnetic field at a distance $R$ from the centre of the magnet on the normal bisector is given by the relation:

$B=\frac{\mu_{0} M}{4 \pi R^{3}}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \,T\,m\, A ^{-1}$

When the resultant field is inclined at $45^{\circ}$ with earth's field, $B=H$

$\therefore \frac{\mu_{0} M}{4 \pi R^{3}}=H=0.42 \times 10^{-4}$

$R^{3}=\frac{\mu_{0} M}{0.42 \times 10^{-4} \times 4 \pi}$

$=\frac{4 \pi \times 10^{-7} \times 5.25 \times 10^{-2}}{4 \pi \times 0.42 \times 10^{-4}}=12.5 \times 10^{-5}$

$\therefore R=0.05\, m =5\, cm$

$(b)$ The magnetic field at a distanced $R^{\prime}$ from the centre of the magnet on its axis is given as

$B^{\prime}=\frac{\mu_{0} 2 M}{4 \pi R^{3}}$

The resultant field is inclined at $45^{\circ}$ with earth's field.

$\therefore B^{\prime}=H$

$\frac{\mu_{0} 2 M}{4 \pi\left(R^{\prime}\right)^{3}}=H$

$\left(R^{\prime}\right)^{3}=\frac{\mu_{0} 2 M}{4 \pi \times H}$

$=\frac{4 \pi \times 10^{-7} \times 2 \times 5.25 \times 10^{-2}}{4 \pi \times 0.42 \times 10^{-4}}=25 \times 10^{-5}$

$R^{\prime}=0.063 \,m =6.3\, cm$