1. Energy stored on an isolated conducting sphere

For an isolated metal sphere of radius $R$ carrying charge $Q$:

$C = 4\pi \varepsilon_0 R$

The electrostatic energy stored is

$U = \frac{Q^2}{2C}$

Substituting $C$:

$U = \frac{Q^2}{2(4\pi \varepsilon_0 R)} = \frac{1}{2}\cdot \frac{1}{4\pi \varepsilon_0}\cdot \frac{Q^2}{R}$

Given,

$\frac{1}{4\pi \varepsilon_0} = 9\times 10^9$

So,

$U = \frac{1}{2}(9\times 10^9)\frac{Q^2}{R}$

2. Insert the given values

Given charge:

$Q = 4\,\mu C = 4\times 10^{-6} C$

Given energy:

$U = 4.5\,J$

Hence,

$4.5 = \frac{1}{2}(9\times 10^9)\frac{(4\times 10^{-6})^2}{R}$

3. Simplify the expression

First,

$(4\times 10^{-6})^2 = 16\times 10^{-12}$

Thus,

$4.5 = \frac{1}{2}(9\times 10^9)\frac{16\times 10^{-12}}{R}$

$4.5 = \frac{1}{2}\cdot 9 \cdot 16 \times 10^{-3} \cdot \frac{1}{R}$

$4.5 = 72\times 10^{-3} \cdot \frac{1}{R}$

$4.5 = \frac{0.072}{R}$

Therefore,

$R = \frac{0.072}{4.5} = 0.016\,m$

4. Convert into mm

$0.016\,m = 16\,mm$

5. Match with the options

The correct option is:

$\boxed{\text{D: }16\,\text{mm}}$