Given:

Galvanometer resistance:

$$ R_g = 30 \, \Omega $$

Full-scale deflection current:

$$ I_g = 20 \, mA = 0.02 \, A $$

Maximum current to be measured:

$$ I = 3 \, A $$

When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ($R_s$) is:

$$ I_s = I - I_g = 3 - 0.02 = 2.98 \, A. $$

Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:

$$ I_g R_g = I_s R_s. $$

Substitute the given values:

$$ 0.02 \times 30 = 2.98 \times R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega. $$

The shunt resistance is represented as:

$$ R_s = \frac{30}{X} \, \Omega. $$

Equate the two expressions:

$$ \frac{30}{X} = 0.2013. $$

Solving for $X$:

$$ X = \frac{30}{0.2013} \approx 149. $$

Thus, the value of $X$ is approximately $$149.$$