To solve this problem, we need to calculate the effective inductance of a circuit consisting of a parallel combination of inductors. We have a series combination of inductors of 4 H and 2 H, and we need to connect an additional inductor in parallel such that the total effective inductance of the circuit is 3 H.

Let's break down the problem step by step:

1. \(L_{\text{series}} = L_1 + L_2 = 4 \, \text{H} + 2 \, \text{H} = 6 \, \text{H}\)

This is the total inductance of the series combination.

Suppose the required inductor that needs to be connected in parallel is \(L_p\). The formula for the total inductance \(L_{\text{total}}\) when two inductances \(L_{\text{series}}\) and \(L_p\) are in parallel is given by:

\(\dfrac{1}{L_{\text{total}}} = \dfrac{1}{L_{\text{series}}} + \dfrac{1}{L_p}\)

Substitute the known values:

\(\dfrac{1}{3} = \dfrac{1}{6} + \dfrac{1}{L_p}\)

Solve for \(\dfrac{1}{L_p}\):

\(\dfrac{1}{L_p} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{2}{6} - \dfrac{1}{6} = \dfrac{1}{6}\)

Thus, the value of \(L_p\) is:

\(L_p = 6 \, \text{H}\)

Therefore, the correct value of the inductor to be connected in parallel for an effective inductance of 3 H is 6 H.

Hence, the correct answer is 6 H.