Two point charges $+10 \mu \mathrm{C}$ and $4 \mu \mathrm{C}$ are placed 10 cm apart in air. The work required to be done to bring them 2 cm closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text { SI units }\right)$

The work done to bring two point charges closer equals the change in their electrostatic potential energy. The potential energy $U$ for point charges $q_1$ and $q_2$ separated by distance $r$ is $U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$.

Given $q_1 = +10 \mu\mathrm{C} = 10 \times 10^{-6}\ \mathrm{C}$, $q_2 = +4 \mu\mathrm{C} = 4 \times 10^{-6}\ \mathrm{C}$, with $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9\ \mathrm{N\ m^2/C^2}$.
Initial separation $r_1 = 0.10\ \mathrm{m}$, final separation $r_2 = 0.08\ \mathrm{m}$ after moving $2\ \mathrm{cm}$ closer.

The work $W$ equals the difference in potential energy: $W = U_2 - U_1$.

Using $U = k \frac{q_1 q_2}{r}$ with $k = 9 \times 10^9$:
$U_1 = \frac{(9 \times 10^9)(10 \times 10^{-6})(4 \times 10^{-6})}{0.10} = 3.6\ \mathrm{J}$
$U_2 = \frac{(9 \times 10^9)(10 \times 10^{-6})(4 \times 10^{-6})}{0.08} = 4.5\ \mathrm{J}$

The work done is $W = 4.5 - 3.6 = 0.9\ \mathrm{J}$.

Alternatively, applying the direct formula:
$W = k q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right) = (9 \times 10^9)(40 \times 10^{-12}) \left( \frac{1}{0.08} - \frac{1}{0.10} \right) = 0.9\ \mathrm{J}$.

Final Answer: $\boxed{0.9\ \mathrm{J}}$