Answer: Option B
$$\eqalign{
& \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr
& = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr
& = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr
& = 5 + 2\sqrt 6 \cr
& \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr
& {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr
& {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr
& = 5 - 2\sqrt 6 \cr
& \therefore {\text{ }}{a^2} + {b^2} \cr
& = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr
& = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr
& = 2\left( {25 + 24} \right) \cr
& = 2 \times 49 \cr
& = 98 \cr} $$