A bar magnet of length 6 ~cm has a magnetic moment of 4 ~J ~T^ -1 ....

Along the equatorial line, magnetic field strength B=$\frac{\mu_0}{4 \pi} \frac{M}{\left(r^2+\ell^2{ }^{3 / 2}\right.}$\right)$Given:$M=4 \mathrm{~J} T^{-1}$r=200$\mathrm{~cm}=$2$\mathrm{~m}\ell=\frac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$\therefore B=$\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{4}{\left[2^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}$Solving we get,$B=5 \times 10^{-8} tesla