Ta có: \({L_A} < {L_B} \Rightarrow OA > OB\).

Gọi I là trung điểm AB. Ta có:

+) \({L_A} = \log \frac{k}{{O{A^2}}} \Rightarrow \frac{k}{{O{A^2}}} = {10^{{L_A}}} \Leftrightarrow OA = \frac{{\sqrt k }}{{\sqrt {{{10}^{{L_A}}}} }}\).

+) \({L_B} = \log \frac{k}{{O{B^2}}} \Rightarrow \frac{k}{{O{B^2}}} = {10^{{L_B}}} \Leftrightarrow OB = \frac{{\sqrt k }}{{\sqrt {{{10}^{{L_B}}}} }}\).

+) \({L_I} = \log \frac{k}{{O{I^2}}} \Rightarrow \frac{k}{{O{I^2}}} = {10^{{L_I}}} \Leftrightarrow OI = \frac{{\sqrt k }}{{\sqrt {{{10}^{{L_I}}}} }}\).

Vì I là trung điểm AB nên ta có: \(OI = \frac{1}{2}\left( {OA - OB} \right) \Leftrightarrow \frac{{\sqrt k }}{{\sqrt {{{10}^{{L_I}}}} }} = \frac{1}{2}\left( {\frac{{\sqrt k }}{{\sqrt {{{10}^{{L_A}}}} }} - \frac{{\sqrt k }}{{\sqrt {{{10}^{{L_B}}}} }}} \right)\)

\( \Leftrightarrow \frac{1}{{\sqrt {{{10}^{{L_I}}}} }} = \frac{1}{2}\left( {\frac{1}{{\sqrt {{{10}^{{L_A}}}} }} - \frac{1}{{\sqrt {{{10}^{{L_B}}}} }}} \right) \Leftrightarrow {L_I} = - 2\log \left[ {\frac{1}{2}\left( {\frac{1}{{\sqrt {{{10}^{{L_A}}}} }} - \frac{1}{{\sqrt {{{10}^{{L_B}}}} }}} \right)} \right]\).

Thay số ta được \({L_I} \approx 5,42\left( B \right)\). Chọn C.