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We have a rectangular solid  surmounted by  the frustum of a pyramid

The volume of the  rectangular solid is easy   =  W * L * H =  8 * 8 * 2  = 128 m^3

The  volume of the  frustum is given by

(height / 3) [ Area  of lower base (AL)  + Area of upper base (AU) +  square  root  (  AL*AU )  ]

height = 4 cm

AL= 8 * 8  = 64  m^2

AU = 2*2 =  4 m^2

So....the volume is

(4/3)  ( 64  + 4  +   sqrt ( 64 * 4) )  =   (4/3) [ 68 +  16]  =  (4/3) ( 84)  =  112 m^3

So  the total volume =   [128 + 112] m^3  =   240 m^3