At a standard sea level conditions  $$\rho _{\infty}=0.002377\;slug/ft^{3} \\\mu _{\infty}=3.737\times 10^{-7}\;slug/ft\cdot sec$$

$$V=120\;mph=120\left ( \frac{88}{60} \right )ft/sec=176\;ft/sec \\q_{\infty}=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}=\frac{1}{2}\times0.002377\times176^{2}=36.8\;lb/ft^{2}$$

$$D=2\;inch=0.167\;ft$$

Reynolds number $$Re=\frac{\rho V D}{\mu}=\frac{0.002377\times187.7\times0.167}{3.737\times10^{-7}} \\=199,382$$

Total frontal surface area of the struts is $$25\left ( 0.167 \right )=4.175\;ft^{2}$$

Drag due to struts , here $C_{D}=1$ $$D_{s}=q_{\infty}SC_{D}=36.8\times4.175\times 1 =153\;lb$$

For the bracing wires $$D=\frac{3}{32}\;inch=0.0078\;ft$$

$$Re=199382\left ( \frac{0.0078}{0.167} \right )=9312$$

For the wires $C_{D}=1$.Total frontal surface area of wires is $$80\times 0.0078=0.624\;ft^{2}$$

Therefore drag due to wires is $$D_{w}=q_{\infty}SC_{D}=36.8\times0.624\times1=23\;lb$$

Total drag due to struts and wires = $D_{s}+D_{w}=153+23=176\;lb$

Total zero-lift drag for the airplane (including struts and wires ) is $$D=q_{\infty}SC_{D,0}=36.8\times230\times0.036=304.8\;lb$$