In a potentiometer arrangement used to compare resistances, the potential drop across a resistance is directly proportional to the balancing length of the wire ($V \propto l$). Assuming the resistors $R$ and $X$ are connected in series such that the same current flows through them, the ratio of resistances is equal to the ratio of their respective balancing lengths: $\frac{X}{R} = \frac{l_X}{l_R}$.

Substituting the given values: $X = R \times \frac{l_X}{l_R} = 10.0 \, \Omega \times \frac{68.5 \text{ cm}}{58.3 \text{ cm}}$ $X \approx 10.0 \times 1.175 = 11.75 \, \Omega$.

The closest value among the options is 11.7 \Omega .