Correct Answer - Option 3 : 2.0
Due to uniformly distributed line load vertical stress
\(= \frac{{2q}}{{\pi z}}{\left[ {\frac{1}{{1 + {{\left( {\frac{x}{z}} \right)}^2}}}} \right]^2}\)
Vertically below line load x = 0
\(\begin{array}{l} {\sigma _z} = \frac{{2q}}{{\pi z}}\\ \frac{{{\sigma _1}}}{{{\sigma _2}}} = \frac{{{z_2}}}{{{z_1}}} = \frac{4}{2} = 2 \end{array}\)