When a current of 10 mA is passed through a tangent galvanometer,...

The formula for a tangent galvanometer is given by:

$B = B_H \tan \theta$

The magnetic field at the center of the coil is:

$B = \dfrac{\mu_0 N I}{2 R}$

Equating the two expressions for $B$:

$\dfrac{\mu_0 N I}{2 R} = B_H \tan \theta$

Rearranging to solve for the number of turns $N$:

$N = \dfrac{2 R B_H \tan \theta}{\mu_0 I}$

Substituting the given values $R = 0.1 \text{ m}$, $B_H = 3.6 \times 10^{-5} \text{ T}$, \theta = 45^{}^{{}^{\circ}}, $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$, and $I = 10 \times 10^{-3} \text{ A}$:

N = \dfrac{2 \times 0.1 \times 3.6 \times 10^{-5} \times \tan 45^{}^{{}^{\circ}}}{4\pi \times 10^{-7} \times 10 \times 10^{-3}}

$N = \dfrac{0.72 \times 10^{-5}}{4\pi \times 10^{-9}}$

$N = \dfrac{7200}{4\pi} = \dfrac{1800}{\pi}$

Using $\pi \approx 3.14$:

$N \approx \dfrac{1800}{3.14} \approx 573$

The closest integer value among the given options is $570$.