The gravitational force acting on a particle of mass $m$ at a distance$r (<R)$ from the centre of Earth is

$F=-G\frac{M^{\text{'}}m}{r^2}$  ....(1)

where $M^{\text{'}}$ is the mass of portion of Earth enclosed by a sphere of radius $r$ concentric with Earth. Here $-$ sign indicates radially inward direction of force.

Taking the density of Earth to be uniform

$\frac{M^{\text{'}}}{4{\mathrm{\pi r}}^3}=\frac{M}{4{\mathrm{\pi R}}^3}$

$\Rightarrow M^{\text{'}}=\frac{M{r}^3}{R^3}$   ....(2)

Substituting $M^{\text{'}}$ from equation 2 into equation 1:

$F=-\left(\frac{GMm}{R^3}\right) r$

Which is similar to the equation of motion $F=-kx$ with $k=G\frac{Mm}{R^3}=\frac{mg}{R} (\because g=G\frac{M}{R^2})$

For such motion, the time period is given by:

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

Substituting k from above,

$T=2\pi \sqrt{\frac{R}{g}}$

Putting the values, we get

$T=2\times 3.14\sqrt{\frac{6400\times {10}^3}{10}}=2\times 3.14\times 800\approx 1 \mathrm{hour} 24 \mathrm{minutes}$