Internal dimensions

$= 4'8'' \text{ by } 3'6'' \text{ by } 4'4''$

External dimensions

$= 4'10'' \text{ by } 3'8'' \text{ by } 4'5''$

Internal volume

$V_i = \left( 4 + \dfrac{8}{12} \right)\left( 3 + \dfrac{6}{12} \right)\left( 4 + \dfrac{4}{12} \right)$

$V_i = \dfrac{637}{9} \, \text{ ft}^3$

External volume

$V_e = \left( 4 + \dfrac{10}{12} \right)\left( 3 + \dfrac{8}{12} \right)\left( 4 + \dfrac{5}{12} \right)$

$V_e = \dfrac{16\,907}{216} \, \text{ ft}^3$

Volume of sheet iron

$V = V_e - V_i$

$V = \dfrac{16\,907}{216} - \dfrac{637}{9}$

$V = \dfrac{1619}{216} \, \text{ ft}^3$

Unit weight of salt water

$\gamma_{sea} = 64 \, \text{ lb/ft}^3$

Unit weight of iron

$\gamma_{iron} = 7.2 \, \gamma_{sea}$

$\gamma_{iron} = 7.2(64)$

$\gamma_{iron} = 460.8 \, \text{ lb/ft}^3$

Weight of tank when empty

$W_{empty} = \gamma_{iron} V$

$W_{empty} = 460.8 \left( \dfrac{1619}{216} \right)$

$W_{empty} = 3453.87 \, \text{ lb}$ *answer*

Weight of tank when full of seawater

$W_{full} = W_{empty} + W_{sea}$

$W_{full} = 3453.87 + \gamma_{sea} V_i$

$W_{full} = 3453.87 + 64(637/9)$

$W_{full} = 7983.64 \, \text{ lb}$ *answer*