Solution :
When the point lies on axial line .
Let the resultant magnetic field $B_r$ makes an angle of $45^{\circ}$
$\therefore B^1 = \large\frac{\mu_0}{4 \pi} \frac{2m}{r^3}$ (S to N)
$\therefore \tan 45^{\circ} = \large\frac{B^1 \sin 90^{\circ}}{B' \cos 90^{\circ}+Be}$
$1= \large\frac{B^1}{B^e} $
$=> B_e =B^1$
$0.42 \times 10^{-4} =\large\frac{\mu_0}{4 \pi } \times \frac{2m}{r^3}$
=> $ 0.42 \times 10^{-4}= \large\frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{r^3}$
$r^3 =\large\frac{10^{-9} \times 2 \times 5.25 }{0.42 \times 10^{-4}}$
$\quad= 2.5 \times 10^{-5}$
$r= (2.5 \times 10^{-5})^{1/3}$
$r= 0.063\;m$ or $ 6.3 \;cm$
Answer : $ 6.3\;cm$