The length of the wire between the pulleys, as shown in the figure,...

Given:
Length of the wire, $L = 1.5 \text{ m}$
Mass of the wire, $m = 12.0 \text{ g} = 12 \times 10^{-3} \text{ kg}$
Mass of the hanging blocks, $M = 9 \text{ kg}$

The linear mass density of the wire is:
$\mu = \dfrac{m}{L} = \dfrac{12 \times 10^{-3}}{1.5} = 8 \times 10^{-3} \text{ kg/m}$

The tension in the wire is provided by the hanging mass (taking $g = 9.8 \text{ m/s}^2$):
$T = Mg = 9 \times 9.8 = 88.2 \text{ N}$

The speed of the transverse wave in the wire is:
$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{88.2}{8 \times 10^{-3}}} = \sqrt{11025} = 105 \text{ m/s}$

Since the wire vibrates in two loops, there is a node at the center. The length of the wire corresponds to two half-wavelengths:
$L = 2 \left(\dfrac{\lambda}{2}\right) \Rightarrow \lambda = L = 1.5 \text{ m}$

The frequency of vibration is:
$f = \dfrac{v}{\lambda} = \dfrac{105}{1.5} = 70 \text{ Hz}$