When you see a square root in a denominator of a fraction, you need to get it out of the denominator - that's a required simplification in math. When that root is in a sum or subtraction, you need to use the difference of squares pattern, as below. Note also we can factor a perfect square from √63, so we can also simplify that root:
\(\begin{align}
\left( 2 \sqrt{63} + \frac{2}{8 + 3\sqrt{7}} \right)^{\frac{1}{2}} &= \left( 2 \sqrt{9} \sqrt{7} + \left( \frac{2}{8 + 3\sqrt{7}} \right) \left( \frac{8 - 3\sqrt{7}}{8 - 3\sqrt{7}} \right) \right)^{\frac{1}{2}} \\
&= \left( 6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{8^2 - (9)(7)} \right)^{\frac{1}{2}} \\
&= \left(6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{1} \right)^{\frac{1}{2}} \\
&= (16)^{\frac{1}{2}} \\
&= 4
\end{align}\)
This is identical to the approach chetan2u used above, but I post it in case the additional detail helps anyone. Estimation also works here, but I wouldn't advise relying on that strategy, since it will only work when the answer choices are sufficiently far apart.
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