1. Given data

• Side of equilateral triangle: $a = 10\text{ cm} = 0.1\text{ m}$
• Magnetic field: $B = 20\text{ mT} = 20 \times 10^{-3}\text{ T}$
• Current: $I = 0.2\text{ A}$

We assume the coil has one turn.

2. Torque on a current loop

The torque on a current-carrying loop is

$\tau = N I A B \sin\theta$

where $\theta$ is the angle between the normal to the plane of the coil and the magnetic field.

When the plane of the coil is parallel to the magnetic field, the normal to the coil is perpendicular to $\vec B$, so

$\theta = 90^\circ \implies \sin\theta = 1$

Hence,

$\tau = IAB$

3. Area of the equilateral triangular coil

Area of an equilateral triangle:

$A = \frac{\sqrt{3}}{4} a^2$

Substitute $a = 0.1\text{ m}$:

$A = \frac{\sqrt{3}}{4}(0.1)^2 = \frac{\sqrt{3}}{4}(0.01)$

$A = 0.0025\sqrt{3}\text{ m}^2$

4. Calculate torque

$\tau = IAB = (0.2)(0.0025\sqrt{3})(20\times10^{-3})$

First multiply the numerical parts:

$0.2 \times 0.0025 = 0.0005$

$0.0005 \times 20\times10^{-3} = 0.0005 \times 0.02 = 10^{-5}$

Therefore,

$\tau = \sqrt{3} \times 10^{-5}\text{ N m}$

5. Compare with the given form

Given,

$\tau = \sqrt{x} \times 10^{-5}\text{ N m}$

So,

$\sqrt{x} = \sqrt{3} \implies x = 3$

6. Final answer

$\boxed{x=3}$

The derived answer matches the stored correct answer.