Đáp án đúng: C
Giải chi tiết:Đặt \({z_1} = x + yi \Rightarrow {z_2} = x - yi\,\,\left( {x;y \in \mathbb{R}} \right)\).Ta có: \(\left| {{z_1} + {z_2}} \right| = 2\sqrt 6  \Leftrightarrow \left| {x + yi + x - yi} \right| = 2\sqrt 6  \Leftrightarrow x =  \pm \sqrt 6 \,\,\left( 1 \right)\).Ta lại có:\(\begin{array}{l}\frac{{{z_1}}}{{z_2^2}} = \frac{{x + yi}}{{{{\left( {x - yi} \right)}^2}}} = \frac{{\left( {x + yi} \right)}}{{\left( {{x^2} - {y^2}} \right) - 2xyi}}\\ = \frac{{\left( {x + yi} \right)\left[ {\left( {{x^2} - {y^2}} \right) + 2xyi} \right]}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}}\\ = \frac{{\left( {x + yi} \right)\left( {{x^2} - {y^2}} \right) + 2\left( {x + yi} \right)xyi}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}}\\ = \frac{{x\left( {{x^2} - {y^2}} \right) + y\left( {{x^2} - {y^2}} \right)i + 2{x^2}yi - 2x{y^2}}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}}\\ = \frac{{x\left( {{x^2} - {y^2}} \right) - 2x{y^2}}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}} + \frac{{y\left( {{x^2} - {y^2}} \right) + 2{x^2}y}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}}i\end{array}\)Vì \(\frac{{{z_1}}}{{z_2^2}}\) là số ảo nên \(\begin{array}{l}\frac{{x\left( {{x^2} - {y^2}} \right) - 2x{y^2}}}{{{{\left( {{x^2} - {y^2}} \right)}^2} + 4{x^2}{y^2}}} = 0 \Leftrightarrow {x^3} - 3x{y^2} = 0\\ \Leftrightarrow x\left( {{x^2} - 3{y^2}} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \sqrt 3 y\\x =  - \sqrt 3 y\end{array} \right.\,\,\left( 2 \right)\end{array}\)Từ (1) và (2) \( \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {\sqrt 6 ;\sqrt 2 } \right);\left( {\sqrt 6 ; - \sqrt 2 } \right);\left( { - \sqrt 6 ;\sqrt 2 } \right);\left( { - \sqrt 6 ; - \sqrt 2 } \right)} \right\}\).Vậy \(\left| {{z_1}} \right| = \sqrt {6 + 2}  = 2\sqrt 2 \).Chọn C