The velocity of the stone at the bottom is just enough to take it to the top of the circle without slackening the string. What is the tension in string when it becomes horizontal
For the stone to just reach the top of the circle
$$\frac 12 mv^2=mg (2R)$$ $$v=\sqrt{4g}$$
When the rock is horizontal $$\frac 12 m4g=mg(R)+\frac 12mv^2$$ $$v=\sqrt{2g}$$
Then $$T=\frac{mv^2}{R}$$ $$T=2g$$ $$T=20N$$ But the answer given is 30 N. what’s wrong?
Your equations are
$$mg=m\dfrac{v^2}{R}$$
for the forces at the top, tension $T$ is $0$! but still veering (some centripetal acceleration exists due to the force of gravity);
$$\dfrac12mv^2+mgR=\dfrac12mV^2$$
for the conservation of energy along the path from the bottom to the position where the string is horizontal and
$$T=m\dfrac{V^2}{R}$$
for the tension at the point considered (no contribution of the force of gravity to the centripetal force.