The energy density ($u$) between the plates of a capacitor is given by the formula:

$$ u = \frac{1}{2} \varepsilon_0 E^2 $$

where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:

$$ E = \frac{V}{d} $$

Given:

Capacitance ($C$) = 1 µF = $1 \times 10^{-6} \, \text{F}$

Potential Difference ($V$) = 20 V

Distance ($d$) = 1 µm = $1 \times 10^{-6} \, \text{m}$

First, calculate the electric field $E$:

$$ E = \frac{V}{d} = \frac{20 \, \text{V}}{1 \times 10^{-6} \, \text{m}} = 2 \times 10^7 \, \text{V/m} $$

Now plug this into the formula for energy density:

$$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2 \times 10^7 \, \text{V/m})^2 $$

Calculate $u$:

$$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14} $$

$$ u = 2 \times 8.85 \times 10^2 $$

$$ u = 1770 \, \text{J/m}^3 $$

This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$. Therefore, the correct option is:

Option A: $1.8 \times 10^3 \, \text{J/m}^3$