Solution

a) According to the definition of electric field strength:
$$F = qE \implies E = \frac{F}{q}$$
In the SI system:
$$E = \frac{1\text{ N}}{1\text{ C}} = 1\text{ V/m}$$
To convert to CGS, we use the relations: $1\text{ V} = \frac{1}{300}\text{ statV}$ and $1\text{ m} = 100\text{ cm}$:
$$E = \frac{1/300\text{ statV}}{100\text{ cm}} = \frac{1}{30000}\text{ statV/cm} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$$

b) For the second case, we first calculate the field strength in the CGS system:
$$E_{\text{cgs}} = \frac{F}{q} = \frac{100\text{ dyn}}{10\text{ statC}} = 10\text{ statV/cm}$$
Now we convert this value to the SI system using the inverse relations ($1\text{ statV} = 300\text{ V}$, $1\text{ cm} = 0.01\text{ m}$):
$$E = 10 \cdot \frac{300\text{ V}}{0.01\text{ m}} = 10 \cdot 30000\text{ V/m} = 3 \cdot 10^5\text{ V/m}$$