$$\lim_{x \to 0} \frac{\int_0^{x^2} \sin\sqrt{ t} \, dt}{x^3}$$ is equal to:

Given,

$$\lim_{x\to0}\frac{\int_0^{x^2}\sin\sqrt t\,dt}{x^3}$$

Put

$$t=u^2$$

Then,

$$dt=2u\,du$$

When

$$t=0,\quad u=0$$

and when

$$t=x^2,\quad u=x$$

Therefore,

$$\int_0^{x^2}\sin\sqrt t\,dt=\int_0^x\sin u\cdot2u\,du$$

Hence limit becomes

$$\lim_{x\to0}\frac{2\int_0^x u\sin u\,du}{x^3}$$

Using

$$\sin u\sim u\quad \text{as }u\to0,$$

we get

$$u\sin u\sim u^2$$

Thus,

$$2\int_0^x u\sin u\,du\sim2\int_0^x u^2\,du$$

$$=2\left[\frac{u^3}{3}\right]_0^x$$

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

Therefore,

$$\lim_{x\to0}\frac{\frac{2x^3}{3}}{x^3}$$

$$=\frac23$$

Hence, the required limit is

$$\boxed{\frac23}$$