Step 1

As per the question:

Thermal conductivity, $k=160\frac{W}{m-K}$

Diameter of the fin, $D=4mm=4\times {10}^{-3}m$

Coefficient of convective heat transfer, $h=220\frac{W}{{m}^{2}-K}$

Efficiency of the fin, ${\eta}_{f}=0.65$

Now,

To calculate the length if the fin:

1) ${\eta}_{f}=\frac{\mathrm{tanh}m{L}_{C}}{m{L}_{C}}$

where

${L}_{C}=L+\frac{D}{4}=(L+1)mm$

Also,

$m=\sqrt{\frac{Ph}{{A}_{C}k}}$

where

$P=\text{Perimeter of the fin}=\pi D$

${A}_{C}=\text{Convection Area}$

${A}_{C}=\frac{\pi {D}^{2}}{4}$

Now,

$m=\sqrt{\frac{\pi Dh}{\frac{\pi {D}^{2}}{4}k}}$

$m=\sqrt{\frac{4h}{Dk}}=\sqrt{\frac{4\times 220}{4\times {10}^{-3}\times 160}}=37.08{m}^{-3}$

Now, using eqn (1)

$0.65=\frac{\mathrm{tanh}37.08{L}_{C}}{37.08{L}_{C}}$

$\mathrm{tanh}(37.08{L}_{C})=24.10{L}_{C}$

Now, by trial-error:

${L}_{C}=0.0362m=36.2mm$

Also, we know that:

${L}_{C}=L+1$

$36.2=L+1$

$L=35.2mm$

Now, the fin effectiveness is given by:

${\epsilon}_{f}=\sqrt{\frac{Pk}{h{A}_{C}}}$

${\epsilon}_{f}=\sqrt{\frac{\pi D\times 160}{220\times \frac{\pi {D}^{2}}{4}}}\mathrm{tanh}m{L}_{C}$

${\epsilon}_{f}\sqrt{\frac{4\times 160}{220\times 4\times {10}^{-3}}}\mathrm{tanh}(37.08\times 0.0362)=23.5$