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Let the volume of the body be $vcm^3$

the of the body be $dgcm^-3$

So weight of the body in air$W_"air"=vdg$ dyne,

$"where "g->"acceleration due to gravity"$

Now the bouyant force on the body when it is completely imerssed is the weight of displaced water

$W_"bouyant"=vxx1xxg$ dyne.

Now the weight of the body in water

$W_"water"=W_"air"-W_"buoyant"=vg(d-1)$ dyne

By the given condition

$W_"water"=1/3W_"air"$

$=>vg(d-1)=1/3vdg$

$=>d-1=1/3d$

$=>d-d/3=1$

$=>(2d)/3=1$

$=>d=3/2=1.5gcm^"-3"$

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