You may solve in this way:

Let \(\sqrt{ 7+ \sqrt{48}} - \sqrt{3} = x\)

\(\sqrt{ 7+ \sqrt{48}} = x + \sqrt{3}\)

Squaring both sides

\(7 + \sqrt{48} = 3 + 2\sqrt{3} x + x^2\)

\(4 + \sqrt{48} = 2\sqrt{3}x + x^2\)

\(x^2 + 2\sqrt{3}x - 43 - 4 = 0\)

\(x^2 - 4 + 2\sqrt{3}(x-2) = 0\)

\((x+2) (x-2) + 2\sqrt{3} (x-2) = 0\)

\((x-2) (x + 2 + 2\sqrt{3}) = 0\)

From the above equation, either x-2 = 0 or \(x + 2 + 2\sqrt{3} = 0\)

x = 2 OR \(x = -(2+2\sqrt{3})\)

From the options, x has a positive value, hence Answer = 2 = C

This is a long method; I would prefer Bunuel's method
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