To determine the emissivity ($\varepsilon$) of the wire used in the bulb, we start with the given parameters and the formula for power radiated by a black body:

Length of the wire ($L$): 10 cm = 0.1 m

Diameter of the wire ($d$): 0.5 mm = 0.0005 m

Temperature of the wire ($T$): $1727^{\circ} \mathrm{C} = 2000 \mathrm{K}$

Power radiated ($P$): 94.2 W

The power radiated by the wire can be expressed as:

$ P = \varepsilon \sigma A T^4 $

Where:

$\sigma$ is the Stefan-Boltzmann constant ($6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$)

$A$ is the surface area of the wire, calculated using the formula for the surface area of a cylinder: $A = \pi d L$

Substituting the given values into the formula:

$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times (3.14 \times 0.0005 \times 0.1) \times (2000)^4 $

Simplifying the expression inside the parentheses:

$ 94.2 = \varepsilon \times (6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4 $

On solving for $\varepsilon$:

$ \varepsilon = \frac{94.2}{[(6 \times 10^{-8}) \times 3.14 \times 0.0005 \times 0.1 \times (2000)^4]} $

Given that $\varepsilon$ is $\frac{x}{8}$, evaluating this expression provides the value of $x = 5$. Thus, the emissivity $\varepsilon$ is $\frac{5}{8}$.