Answer
$7.4\times 10^{-3}J$
Work Step by Step
The required energy can be calculated as follows:
$Dose \space in\space rad=\frac{dose \space in \space rem}{RBE}$
$Dose \space in \space rad=(\frac{35\times 10^{-3}rem}{0.85})(\frac{0.01J/Kg}{rem})=0.41176\times 10^{-3}J/Kg$
The energy absorbed is
$E=(0.41176\times 10^{-3}J/Kg)(72Kg)=29.647\times 10^{-3}J$
and the energy absorbed by $\frac{1}{4}th$ of the patient's body is given as
$\frac{E}{4}=\frac{29.647\times 10^{-3}J}{4}=7.4\times 10^{-3}J$
