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
$W_"water"=1/3W_"air"$
$=>vg(d-1)=1/3vdg$
$=>d-1=1/3d$
$=>d-d/3=1$
$=>(2d)/3=1$
$=>d=3/2=1.5gcm^"-3"$