Answer
$0.091^{\circ}$
Work Step by Step
We know that the,
$F=S(\frac{\Delta X}{L_{0}})A=>(\frac{\Delta X}{L_{0}})=\frac{F}{SA}-(1)$
From the figure, we can write,
$tan\theta=(\frac{\Delta X}{L_{0}})=>\theta=tan^{-1}(\frac{\Delta X}{L_{0}})-(2)$
(1)=>(2),
$\theta=tan^{-1}[\frac{6\times10^{6}N}{(4.2\times10^{10}N/m^{2})(0.09\space m^{2})}]\approx0.091^{\circ}$
Here the shear modulus S for copper is taken from table 10.2
