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(A) In $	riangle AEG$,$\frac{EG}{AG} = 	an 30^{\circ}$. Since $AG = 10 \, cm$,$EG = 10 	an 30^{\circ} = \frac{10}{\sqrt{3}} \, cm$. This is the rad

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$7964.44$

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(A) In $	riangle AEG$,$\frac{EG}{AG} = 	an 30^{\circ}$. Since $AG = 10 \, cm$,$EG = 10 	an 30^{\circ} = \frac{10}{\sqrt{3}} \, cm$. This is the radius $r_1$ of the top of the frustum.In $	riangle ABD$,$\frac{BD}{AD} = 	an 30^{\circ}$. Since $AD = 20 \, cm$,$BD = 20 	an 30^{\circ} = \frac{20}{\sqrt{3}} \, cm$. This is the radius $r_2$ of the base of the frustum.The height $h$ of the frustum is $10 \, cm$.The volume of the frustum is $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.$V = \frac{1}{3} 	imes \frac{22}{7} 	imes 10 \left[ \left( \frac{10}{\sqrt{3}} ight)^2 + \left( \frac{20}{\sqrt{3}} ight)^2 + \left( \frac{10}{\sqrt{3}} 	imes \frac{20}{\sqrt{3}} ight) ight]$$V = \frac{220}{21} \left[ \frac{100}{3} + \frac{400}{3} + \frac{200}{3} ight] = \frac{220}{21} 	imes \frac{700}{3} = \frac{22000}{9} \, cm^3$.The wire has a radius $r = \frac{1}{32} \, cm$. Let its length be $l$.Volume of wire = $\pi r^2 l = \frac{22}{7} 	imes \left( \frac{1}{32} ight)^2 	imes l$.Equating volumes: $\frac{22000}{9} = \frac{22}{7} 	imes \frac{1}{1024} 	imes l$.$l = \frac{22000}{9} 	imes \frac{7 	imes 1024}{22} = \frac{1000 	imes 7168}{9} = \frac{7168000}{9} \approx 796444.44 \, cm$.Converting to meters: $l = 7964.44 \, m$.

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