Just do it

Suppose the extension in the spring is $x$ .

Three forces are acting on the block--

(1) $mg$ (2) Force due to extension in the spring = $kx$ (3) Normal reaction form the hemisphere.

As the work done by last two forces are zero so we can use conservation of energy to find the velocity of the block at the required instant. Let that velocity be $v$ .

So $mgR(1-cos(60))=0.5mv^{2}$ .....(1)

When it leaves contact with the hemisphere centripetal force would be provided by force due to spring and a component of $mg$ .

So $mv^{2}/R =kx+mgcos(60)$ .......(2)

Using 1) and 2) we get $x=0.5$ So natural length of the spring is $1cm$ .

Let's think why will it leave the surface. Viewing from surface frame 1. centrifugal force would be acting radially outwards. 2. Mgcos(theta) radially inwards. 3. Kx radially inwards. When centrifugal force will become greater than sum of other two forces it will leave the surface. ........ (¡) From energy conservation- Change in KE when it makes 60° with vertical=change in PE. (1/2) m v^2=mg(r/2) (mv^2)/r=mg ...(ii) From (i)... (mv^2)/r=mgcos(60°)+kx Putting value From (ii).. mg=mg/2+kx (mg)/(2k)=x Unstretched length = r-x