Percent increased in volume

let $\color{#D61F06}\large v$ be the volume of the original cube and $\color{#D61F06}\large V$ be the volume of the bigger cube

Since the two cubes are similar, we have

$\color{#D61F06}\large \dfrac{v}{V}=\dfrac{x^3}{(1.6x)^3}=\dfrac{x^3}{4.096x^3}=\dfrac{1}{4.096}$ $\large \implies$ $\large \color{#D61F06} V=4.096v$

Therefore,

$\color{#D61F06}\large \%~increased~in~volume=(4.096-1)(100\%)=$ $\boxed{\color{#D61F06}\large 309.6\%}$