Answer: Option B
$$\eqalign{
& {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr
& \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr
& \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr
& {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr
& \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr
& \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr
& \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr
& \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr
& \,\,\,\,\,\,\, = \frac{{28}}{4} \cr
& \,\,\,\,\,\,\, = 7 \cr
& {\text{Also, }} \cr
& ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr
& \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr
& \,\,\,\,\,\,\, = 1 \cr
& \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr
& \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr
& \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr
& \,\,\,\,\,\,\, = \frac{8}{6} \cr
& \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$