An unknown resistance wire in the left gap of a meter bridge...

Let the unknown resistance in the left gap be $R$ and the resistance in the right gap be $S = 10\ \Omega$.

At the balance point of the meter bridge, the ratio of resistances is equal to the ratio of the lengths of the bridge wire segments:

$\dfrac{R}{S} = \dfrac{l_1}{l_2}$

Given that the balance point divides the bridge wire in the ratio $3:2$, we have:

$\dfrac{R}{10} = \dfrac{3}{2}$

$R = 10 \times \dfrac{3}{2} = 15\ \Omega$

The length of the unknown resistance wire is $L = 1.5\ \text{m}$.

The resistance per unit length of the wire is:

$\dfrac{R}{L} = \dfrac{15}{1.5} = 10.0\ \Omega/\text{m}$

Answer: $10.0\ \Omega/\text{m}$