Given : $$x=\frac{1}{\sqrt{3}}$$ and $$y=\frac{1}{\sqrt{5}}$$

=> $$x^2=\frac{1}{3}$$ and $$y^2=\frac{1}{5}$$

Again, squaring both terms, we get : $$x^4=\frac{1}{9}$$ and $$y^4=\frac{1}{25}$$ -------------(i)

To find : $$(6x^2 - 5y^2)(6x^2 + 5y^2)$$

Using, $$(a-b)(a+b)=a^2-b^2$$ and substituting values from equation (i),

= $$36x^4-25y^4$$

= $$(36\times\frac{1}{9})-(25\times\frac{1}{25})$$

= $$4-1=3$$

=> Ans - (B)