Let r_{1} cm and r_{2} cm denote the radii of the base of the cylinder and cone respectively. Then,

r_{1} = r_{2} = 8 cm

Let h_{1} and h_{2} cm be the height of the cylinder and the cone respectively. Then

h_{1} = 240 cm and h_{2} = 36 cm

Now, volume of the cylinder = πr_{1}^{2}h_{1} cm^{3}

= (π × 8 × 8 × 240)cm^{3}

= (π × 64 × 240) cm^{3}

Volume of the cone = 1/3πr_{2}^{2}h_{2} cm^{3}

= (1/3π × 8 × 8 × 36) cm^{3}

= (1/3π × 64 × 36) cm^{3}

∴ Total volume of iron = Volume of the cylinder + Volume of the cone

= (π × 64 × 240 + 1/3π × 64 × 36) cm^{3}

= π × 64 × (240 + 12) cm^{3}

= 22/7 × 64 × 252 cm^{3}

= 22 × 64 × 36 cm^{3}

Hence, total weight of the pillar = Volume Weight per cm^{3}

= (22 × 64 × 36) × 7.8 gms

= 395366.4 gms

= 395.3664 kg.