${A_n} = \left( {\frac{3}{4}} \right) – {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} – …… + {\left( { – 1} \right)^{n – 1}}{\left( {\frac{3}{4}} \right)^n}$

Which is a $G.P.$ with $a = \frac{3}{4}'r = \frac{{ – 3}}{4}$ and number of terms $=n$

${A_n} = \frac{{\frac{3}{4} \times \left( {1 – {{\left( {\frac{{ – 3}}{4}} \right)}^n}} \right)}}{{1 – \left( {\frac{{ – 3}}{4}} \right)}} – \frac{{\frac{3}{4} \times \left( {1 – {{\left( {\frac{{ – 3}}{4}} \right)}^n}} \right)}}{{\frac{7}{4}}}$

$ \Rightarrow {A_n} = \frac{3}{7}\left[ {1 – {{\left( {\frac{{ – 3}}{4}} \right)}^n}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)$

As, ${B_n} = 1 – {A_n}$

For least odd natural number $p$, such that ${B_n} > {A_n}$

$ \Rightarrow 1 – {A_n} > {A_n}\,\,\,\,\, \Rightarrow 1 > 2 >  \times {A_n}\,\,\,\,\, \Rightarrow {A_n} < \frac{1}{2}$

From eqn. $(1)$, we get 

$\frac{3}{7} \times \left[ {1 – {{\left( {\frac{{ – 3}}{4}} \right)}^n}} \right] < \frac{1}{2}\,\, \Rightarrow 1 – {\left( {\frac{{ – 3}}{4}} \right)^n} < \frac{7}{6}$

$ \Rightarrow 1 – \frac{7}{6} < {\left( {\frac{{ – 3}}{4}} \right)^n} \Rightarrow \frac{{ – 1}}{6} < {\left( {\frac{{ – 3}}{4}} \right)^n}$

As $n$ is odd, then ${\left( {\frac{{ – 3}}{4}} \right)^n} =  – \frac{{{3^n}}}{4}$

So $\frac{{ – 1}}{6} <  – {\left( {\frac{3}{4}} \right)^n}\,\,\, \Rightarrow \frac{1}{6} > {\left( {\frac{3}{4}} \right)^n}$

$\log \left( {\frac{1}{6}} \right) = n\,\log \left( {\frac{3}{4}} \right) \Rightarrow 6.228 < n$

Hence, $n$ should be $7$.