Given displacement equation

The SHM is: $x(t)=A\sin\left(\frac{\pi t}{90}\right)$

Comparing with the standard form $x=A\sin(\omega t),$ we get $\omega=\frac{\pi}{90}$

Find displacement at $t=210\,\text{s}$

$x(210)=A\sin\left(\frac{\pi\cdot 210}{90}\right)$

$x(210)=A\sin\left(\frac{7\pi}{3}\right)$

Since $\sin\left(\frac{7\pi}{3}\right)=\sin\left(2\pi+\frac{\pi}{3}\right)=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2},$

therefore $x=\frac{\sqrt{3}}{2}A$

Use energy relations in SHM

Total energy: $E=\frac{1}{2}kA^2$

Potential energy at displacement $x$: $U=\frac{1}{2}kx^2$

Kinetic energy: $K=E-U=\frac{1}{2}k(A^2-x^2)$

Hence, $\frac{K}{U}=\frac{A^2-x^2}{x^2}$

Substitute $x=\frac{\sqrt{3}}{2}A$

$x^2=\frac{3A^2}{4}$

So, $\frac{K}{U}=\frac{A^2-\frac{3A^2}{4}}{\frac{3A^2}{4}}$

$=\frac{\frac{A^2}{4}}{\frac{3A^2}{4}}$

$=\frac{1}{3}$

Check options

The required ratio is $\boxed{\frac{1}{3}}$

But this value is not present in the options.

Comparison with stored answer

Stored correct answer is B: 3.

However, our derived answer is $\frac{K}{U}=\frac{1}{3}$

So the stored answer appears to be incorrect. It is possible that the question intended the ratio of potential to kinetic energy, for which $\frac{U}{K}=3.$