Assume the length of the pipe be h cm.

Formula for volume of cuboid is $V=whl$

Now, Volume of cuboid $=\left( 53\times 40\times 15 \right)c{{m}^{3}}$

Internal radius of the pipe $=7/2cm=r$

External radius of the pipe $=8/2=4cm=R$

Therefore, the volume of iron in the pipe $=$ (External Volume) $–$ (Internal Volume)

$=\pi {{R}^{2}}h-\pi {{r}^{2}}h$

$=\pi h\left( {{R}^{2}}-{{r}^{2}} \right)$

$=\pi h\left( R-r \right)\left( R+r \right)$

$=\pi \left( 4-7/2 \right)\left( 4+7/2 \right)\times h$

$=\pi \left( 1/2 \right)\left( 15/2 \right)\times h$

Now from the question it’s understood that,

The volume of iron in the pipe $=$ volume of iron in cuboid

$\pi \left( 1/2 \right)\left( 15/2 \right)\times h=53\times 40\times 15$

$h=\left( 53\times 40\times 15\times 7/22\times 2/15\times 2 \right)cm$

$h=2698cm$

Hence, the length of the pipe is $2698cm$.