The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 $$\mu$$C charge, its radius will be: $$\left[\text{Take } : \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}\right]$$

We are told that the energy stored in the electric field of a charged metal sphere is $$U = 4.5\ \text{J}$$ and that the charge on the sphere is $$q = 4\ \mu\text{C}$$.

First we convert the charge into SI units. We have

$$4\ \mu\text{C} = 4 \times 10^{-6}\ \text{C}.$$

The electrostatic energy stored in an isolated conductor of capacitance $$C$$ carrying charge $$q$$ is given by the standard formula

$$U = \frac{q^2}{2\,C}.$$

For a spherical conductor of radius $$R$$, the capacitance is known to be

$$C = 4\pi\epsilon_0 R.$$

Substituting this value of $$C$$ into the energy formula, we get

$$U = \frac{q^2}{2 \times 4\pi\epsilon_0 R}.$$

Simplifying the denominator,

$$U = \frac{q^2}{8\pi\epsilon_0 R}.$$

We now solve this equation for the unknown radius $$R$$. Rearranging,

$$R = \frac{q^2}{8\pi\epsilon_0 U}.$$

The constant $$\dfrac{1}{4\pi\epsilon_0}$$ is given in the question as

$$\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \text{N m}^2\text{ C}^{-2}.$$

Hence,

$$\frac{1}{8\pi\epsilon_0} \;=\; \frac{1}{2}\,\frac{1}{4\pi\epsilon_0} \;=\; \frac{1}{2}\left(9 \times 10^{9}\right) \;=\; 4.5 \times 10^{9}\ \text{N m}^2\text{ C}^{-2}.$$

Putting the numerical values into our expression for $$R$$, we have

$$R \;=\; \left(4.5 \times 10^{9}\right) \frac{q^2}{U}.$$

Substituting $$q = 4 \times 10^{-6}\ \text{C}$$ and $$U = 4.5\ \text{J},$$

$$q^2 = \left(4 \times 10^{-6}\right)^2 = 16 \times 10^{-12}\ \text{C}^2 = 1.6 \times 10^{-11}\ \text{C}^2.$$

Therefore,

$$R = \left(4.5 \times 10^{9}\right)\frac{1.6 \times 10^{-11}}{4.5}.$$

We notice that the factor of $$4.5$$ in the numerator and denominator cancels, leaving

$$R = 1.6 \times 10^{-11} \times 10^{9} = 1.6 \times 10^{-2}\ \text{m}.$$

In decimal form,

$$R = 0.016\ \text{m}.$$

Converting metres to millimetres ($$1\ \text{m} = 1000\ \text{mm}$$),

$$R = 0.016 \times 1000\ \text{mm} = 16\ \text{mm}.$$

Hence, the correct answer is Option B.