(B) We use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.Here,$a = 7$ and $b = 3 \sqrt{5}$.So,$(7+3 \sqrt{5})(7-3 \sqrt{5}) = 7^2 - (3 \sqrt{5})^2$.

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(B) We use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.Here,$a = 7$ and $b = 3 \sqrt{5}$.So,$(7+3 \sqrt{5})(7-3 \sqrt{5}) = 7^2 - (3 \sqrt{5})^2$.Calculating the squares: $7^2 = 49$ and $(3 \sqrt{5})^2 = 9 \times 5 = 45$.Thus,$49 - 45 = 4$.The square root of the result is $\sqrt{4} = 2$.