Let in a Binomial distribution, consisting of 5 independent trials,...

$P\left(X = 1\right) = C1 5 \cdot p . {q}^{4} = 0 . 4096$

$P\left(X = 2\right) = C2 5 \cdot {p}^{2} \cdot {q}^{3} = 0 . 2048$

$⇒ \dfrac{q}{2p} = 2$

$⇒ q = 4p$ and $p + q = 1$

$⇒ p = \dfrac{1}{5}$ and $q = \dfrac{4}{5}$

Now

$P\left(X = 3\right) = C3 5 \cdot {\left(\dfrac{1}{5}\right)}^{3} \cdot {\left(\dfrac{4}{5}\right)}^{2} = \dfrac{10 \times 16}{125 \times 25} = \dfrac{32}{625}$