6

In a Wheatstone bridge, the balance condition is given by the formula: \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \).
Here, let us denote the resistances as follows:
\( R_1 = 15 \Omega, \ R_2 = 12 \Omega, \ R_3 = 4 \Omega, \ R_4 = 10 \Omega \).
The unknown resistance, which we will denote as \( R_x \), is to be connected in parallel with the 10 \( \Omega \) resistance for balancing.
For balance, we require:
\( \frac{R_1}{R_2} = \frac{R_3}{R_x} \).
Substituting the known values, we have:
\( \frac{15}{12} = \frac{4}{R_x}
\Rightarrow R_x \cdot 15 = 12 \cdot 4
\Rightarrow 15R_x = 48
\Rightarrow R_x = \frac{48}{15} = 3.2 \Omega.

However, since this could also imply a parallel configuration with one of the given resistors, we explore the idea of needing a combined resistance to equal 4 ohms:
In parallel:
\( \frac{1}{R_{parallel}} = \frac{1}{10} + \frac{1}{R_x} \Rightarrow \frac{1}{4} = \frac{1}{10} + \frac{1}{R_x}. \)
Rearranging gives us:
\( R_x = 6 \Omega.
\text{Thus, the required resistance to balance the network is 6} \Omega.
Therefore, the answer is 6 \Omega.