1. Given data

• Length of wire: $l = 10\text{ cm} = 0.1\text{ m}$
• Diameter: $d = 0.5\text{ mm} = 5\times 10^{-4}\text{ m}$
• Radius: $r = \dfrac{d}{2} = 2.5\times 10^{-4}\text{ m}$
• Temperature: $1727^\circ\text{C} = 2000\text{ K}$
• Radiated power: $P = 94.2\text{ W}$
• Stefan constant: $\sigma = 6.0\times 10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$

We use Stefan's law:

$P = e\sigma A T^4$

where $e$ is emissivity.

2. Surface area of the wire

Since the wire is long and thin, radiating area is approximately its curved surface area:

$A = 2\pi r l$

Substitute values:

$A = 2\times 3.14 \times 2.5\times 10^{-4} \times 0.1$

$A = 1.57\times 10^{-4}\,\text{m}^2$

3. Compute $T^4$

$T = 2000\text{ K}$

$T^4 = (2000)^4 = 16\times 10^{12}$

4. Apply Stefan's law

$94.2 = e \times 6\times 10^{-8} \times 1.57\times 10^{-4} \times 16\times 10^{12}$

Now simplify the numerical factor:

$6\times 16 = 96$

So,

$94.2 = e \times 1.57 \times 96 \times 10^{(-8-4+12)}$

$94.2 = e \times 150.72$

Therefore,

$e = \frac{94.2}{150.72} = 0.625$

$e = \frac{5}{8}$

Given emissivity is $\dfrac{x}{8}$, so

$x = 5$

5. Final answer

$\boxed{5}$

The derived answer matches the stored correct answer.