Given, the magnetic moment of the bar magnet $M=0.48 \mathrm{J} {\mathrm{T}}^{-1}$

Distance from the centre of the magnet $d=10 \mathrm{cm}=0.1 \mathrm{m}$

When the point lies on the axial line. The magnetic field at point P

          

                  $\text{B}=\frac{{\mu }_0}{4\pi }·\frac{2\text{M}}{{\text{d}}^3}$

                     $=\frac{10^{-7}\times 2\times 0\text{.}48}{{\left(0\text{.}1\right)}^3}=0\text{.}96\times 10^{-4}\text{ T}$

The direction of the magnetic field is along the direction of the magnetic moment. We know that the direction of magnetic moment is from S to N pole. Thus, the direction of the magnetic field is from S to N pole of the magnet.