Given $F=-k{x}^3$

$-\frac{dU}{dx}=-k{x}^3$

$\Rightarrow U=\frac{1}{4}k{x}^4$

$\therefore$ Energy of oscillations will be

$E=\frac{1}{2}m{v}^2+U=\frac{1}{2} m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{4}k{x}^4\dots \left(1\right)$

If we pull $\frac{dx}{dt}=0$ in above equation, we will

get amplitude as $\mathrm{A}=\sqrt[4]{\frac{4\mathrm{E}}{\mathrm{k}}}\dots \left(2\right)$

Also on rearranging equation $\left(1\right),$ we get

$dt=\pm dx\sqrt{\frac{m}{2E}}{\left(1-\frac{k}{4E}{x}^4\right)}^{-1/2}$

Now, use $A=\sqrt[4]{\frac{4E}{k}},$ to reduce above equation

as $dt=\pm dx\sqrt{\frac{2 \mathrm{m}}{k}}{A}^{-2}{\left(1-{\left(\frac{x}{A}\right)}^4\right)}^{-1/2}$

The time period can be found by integrating above equation.

$T=4{\int }_0^{A}dx\sqrt{\frac{2 \mathrm{m}}{k}}{A}^{-2}{\left(1-{\left(\frac{x}{A}\right)}^4\right)}^{-1/2}$

$=4\sqrt{\frac{2\mathrm{m}}{k}}{A}^{-2}{\int }_0^A{\left(1-{\left(\frac{x}{A}\right)}^4\right)}^{-1/2}·dx$

Put $\frac{x}{A}=u\Rightarrow dx=Adu$

$\therefore T=4\sqrt{\frac{2 \mathrm{m}}{k}}{A}^{-2}( A){\int }_0^{1}du{\left(1-u^4\right)}^{-1/2}$

$T=4\sqrt{\frac{2 \mathrm{m}}{k}}{A}^{-1}\left(\mathrm{I}\right)$

Where $I={\int }_0^1{\left(1-u^4\right)}^{-1/2}du$ is a numerical value

So from above equation $T\propto A^{-1}$

$\therefore \frac{T_1}{ T_2}=\frac{2 A}{ A}\Rightarrow T_2=\frac{T}{2}$