Correct Answer: $4-2 \sqrt{2}$

Solution : This is a telescoping series, where each term cancels out a part of the next term.
$\frac{1}{4-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+\frac{1}{\sqrt{14}-\sqrt{13}}-\frac{1}{\sqrt{13}-\sqrt{12}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{10}-3}-\frac{1}{3-\sqrt{8}}$
Rationalising the denominator and use the formula $a^2 - b^2 = (a+b)(a-b)$, we get,
$=\frac{\sqrt{15}+4}{1}-\frac{\sqrt{15}+\sqrt{14}}{1}+\frac{\sqrt{14}+\sqrt{13}}{1}-\frac{\sqrt{13}+\sqrt{12}}{1}+\frac{\sqrt{12}+\sqrt{11}}{1}-\frac{\sqrt{11}+\sqrt{10}}{1}+\frac{\sqrt{10}+3}{1}-\frac{3+\sqrt{8}}{1}$
$=\sqrt{15}+4-\sqrt{15}-\sqrt{14}+\sqrt{14}+\sqrt{13}-\sqrt{13}-\sqrt{12}+\sqrt{12}+\sqrt{11}-\sqrt{11}-\sqrt{10}+\sqrt{10}+3-3-\sqrt{8}$
$=4-\sqrt{8}$
$=4-2\sqrt{2}$
Hence, the correct answer is $4-2\sqrt{2}$.