Answer
$\theta=28.1^o$
Work Step by Step
$\theta$ is the angle the string makes with the vertical. From this angle, the tension force $T$ in the string will have 2 components: horizontal one $T\sin\theta$ and vertical one $T\cos\theta$
After the block swings toward the outside of the curve, it makes an angle $\theta$ and stays put there. That means it does not have acceleration, and all net forces on the block equal $0$, meaning all forces balance one another.
- Horizontal: $T\sin\theta$ opposes the centripetal force $F_c$, so $$T\sin\theta=F_c=\frac{mv^2}{r} (1)$$
- Vertical: $T\cos\theta$ opposes the block's weight $mg$, so $$T\cos\theta=mg (2)$$
From (1) and (2), we have $$\frac{T\sin\theta}{T\cos\theta}=\frac{\frac{mv^2}{r}}{mg}$$ $$\tan\theta=\frac{v^2}{rg}$$
We have $v=28m/s$, $r=150m$ and $g=9.8m/s^2$, so $$\tan\theta=0.533$$ $$\theta=28.1^o$$
