To solve this problem it is necessary to apply the definition of Young's Module which states that
[tex]Y_1 = \frac{\frac{F}{A}}{\frac{\Delta l_0}{l}}[/tex]
Where,
F = Force
A = Cross sectional Area
L = Length
[tex]L_0[/tex] = Initial Length
We need to find the ratio between the two values when the another values are constant, that is
[tex]\frac{Y_1}{Y_2} = \frac{\frac{\frac{F}{A}}{\frac{\Delta l_1}{l}}}{\frac{\frac{F}{A}}{\frac{\Delta l_2}{l}}}[/tex]
[tex]\frac{Y_1}{Y_2} = \frac{\Delta l_2}{\Delta l_1}[/tex]
Re-arrange to find [tex]\Delta l_2,[/tex]
[tex]\Delta l_2 = \frac{9.4*10^9}{1.6*10^{10}}*3.7*10^{-5}[/tex]
[tex]\Delta l_2 = 2.17*10^{-5} m[/tex]
Therefore the bone stretch around [tex] 2.17*10^{-5} m[/tex]