Official solution
A(\[\lambda \]) = \[2\int\limits_0^\lambda {\sqrt x } dx\]
= \[2\left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^\lambda \]
= \[{4 \over 3}{\lambda ^{{3 \over 2}}}\]
\[ \therefore \] A(4) = \[{4 \over 3}{\left( 4 \right)^{{3 \over 2}}}\]
Given, \[{{A\left( \lambda \right)} \over {A\left( 4 \right)}} = {2 \over 5}\]
\[ \Rightarrow \] \[{{{4 \over 3}{{\left( \lambda \right)}^{{3 \over 2}}}} \over {{4 \over 3}{{\left( 4 \right)}^{{3 \over 2}}}}} = {2 \over 5}\]
\[ \Rightarrow \] \[{\lambda ^{{3 \over 2}}} = {2 \over 5} \times 8\]
\[ \Rightarrow \] \[\lambda \] = \[{\left( {{{16} \over 5}} \right)^{{2 \over 3}}}\]
\[ \Rightarrow \] \[\lambda \] = \[{\left( {{{256} \over {25}}} \right)^{{1 \over 3}}}\]
\[ \Rightarrow \] \[\lambda \] = \[{\left( {{{{4^3} \times 4} \over {25}}} \right)^{{1 \over 3}}}\]
= \[4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}\]