When an ammeter is designed, a shunt resistor ($R_s$) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself. This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.

In this scenario, we are told that $5\%$ of the main current passes through the galvanometer. This means that the remaining $95\%$ of the current must pass through the shunt. Let's denote the total current as $I$, the current through the galvanometer as $I_g$, and the current through the shunt as $I_s$. Therefore, we have:

$$ I_g = \frac{5}{100} I $$

$$ I_s = I - I_g = I - \frac{5}{100} I = \frac{95}{100} I $$

Since the galvanometer and shunt are in parallel, the voltage across each must be the same:

$$ V_g = V_s $$

According to Ohm's law, $V = IR$, where $V$ is the voltage, $I$ is the current, and $R$ is the resistance. Hence, for the galvanometer and the shunt:

$$ I_g G = I_s R_s $$

By substituting $I_g$ and $I_s$ from the above proportionality, we get:

$$ \left(\frac{5}{100} I\right) G = \left(\frac{95}{100} I\right) R_s $$

Let's solve for $R_s$:

$$ R_s = \frac{5}{95} G $$

Further simplifying this:

$$ R_s = \frac{G}{19} $$

The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:

$$ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s} $$

Substitute $R_s$ with $\frac{G}{19}$:

$$ \frac{1}{R_a} = \frac{1}{G} + \frac{19}{G} $$

$$ \frac{1}{R_a} = \frac{20}{G} $$

Thus the resistance of the ammeter $R_a$ is:

$$ R_a = \frac{G}{20} $$

Hence, the correct answer is Option C: $\frac{G}{20}$.