Solution

1. In the SI system:
From Coulomb's law:
$$F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{r^2}$$
Expressing the charge $q$:
$$q = \sqrt{4\pi\varepsilon_0 r^2 F}$$
Substitute the numerical values ($F = 1\text{ N}$, $r = 1\text{ m}$, $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$):
$$q = \sqrt{\frac{1 \cdot 1^2}{9 \cdot 10^9}} = \frac{1}{3 \cdot 10^4 \cdot \sqrt{10}} \approx 1.05 \cdot 10^{-5}\text{ C}$$

2. In the CGS system (electrostatic):
In the CGS system, the coefficient is $1$, and Coulomb's law is written as:
$$F = \frac{q^2}{r^2}$$
Expressing the charge $q_{\text{cgs}}$:
$$q_{\text{cgs}} = r\sqrt{F}$$
Keep in mind that in CGS, force is measured in dynes ($1\text{ N} = 10^5\text{ dyn}$), and distance is measured in centimeters ($1\text{ m} = 100\text{ cm}$).
$$q_{\text{cgs}} = 100\text{ cm} \cdot \sqrt{10^5\text{ dyn}} = 100 \cdot 316.2 = 3.16 \cdot 10^4\text{ statC (esu)}$$